## Sir Andrew Wiles on being bright at math:

Yes, some people are brighter than others in math. But I really believe that most people can really reach a good level in math if they are prepared to handle pyschological issues arising out of situations when being stuck.

## Derivatives part 14: IITJEE maths tutorial problems for practice

This is part 14 of the series

Question 1:

Let $f(x)= \sqrt{x-1} + \sqrt{x+24-10\sqrt{x-1}}$ for $x<26$ be a real valued function. Then, find $f^{'}(x)$ for $1:

Consider $(\sqrt{x-1}-5)^{2} = x-1+25-10\sqrt{x-1} = x+24 -10\sqrt{x-1}$ so that we have

$\sqrt{x+24-10\sqrt{x-1}}=\sqrt{x-1}-5$

Hence, $f(x) = \sqrt{x-1} + \sqrt{x-1}-5 = 2\sqrt{x-1}-5$ when $1

Hence, $f^{'}{x} = \frac{-2}{2\sqrt{x-1}} = -\frac{1}{\sqrt{x-1}}$

Question 2:

Let $3f(x) - 2 f(\frac{1}{x})=x$, then find $f^{'}(2)$.

Given that $3f(x) - 2 f(\frac{1}{x})=x$….call this I.

Also, from above, we get $3f(\frac{1}{x}) - 2 f(x)= \frac{1}{x}$…call this II.

so we get $6f(x)-4f(\frac{1}{x})=2x$….call this I’

and $9f(\frac{1}{x})-6f(x) = 9/x$…call this II’.

$5f(\frac{1}{x})=2x+ \frac{9}{x}$ and hence, $f^{'}(1/x) = \frac{2x}{5} + \frac{9}{5x}$

Also, again $3f(x)-2f(1/x)=x$….A

$3f(1/x)-2f(x)=1/x$…B

So, we now we get the following two equations:

$9f(x)-6f(1/x)=3x$…..A’

$6f(1/x)-4f(x)=2/x$….B’

so, now we have $5f(x) = 3x + \frac{2}{x}$ so that we get $f(x) = \frac{3}{5}x+\frac{2}{5x}$ and$f(1/x) = \frac{2}{5}x+\frac{9}{5x}$

so $f^{'}(x) = \frac{3}{5} + \frac{2}{5}\frac{-1}{x^{2}} = \frac{3}{5} - \frac{1}{10}=\frac{1}{2}$

Question 3:

If $x= \frac{a(1-t^{2})}{1+t^{2}}$ and $y = \frac{2bt}{1+t^{2}}$, then find $\frac{dy}{dx}$.

Given that $x = \frac{a(1-t^{2})}{1+t^{2}}$ where a is a parameter (constant) and t is a variable.

Let $t=\tan{\theta}$ so that $x = \frac{a(1-\tan^{2}{\theta})}{1+\tan^{2}{\theta}} = a \cos{2\theta}$

so that $y = \frac{2bt}{1+t^{2}}=b \sin{2\theta}$

$\frac{dy}{dx} = \frac{\frac{dy}{d\theta}}{\frac{dx}{d\theta}} = \frac{2b\cos{2\theta}}{-2a\sin{2\theta}}=- \frac{b}{a\tan{2\theta}}$

so that we have

$\frac{dy}{dx} = - \frac{b(1-t^{2})}{a}$

Question 4:

If $y = \arccos{\frac{x-x^{-1}}{x+x^{-1}}}$ then find $\frac{dy}{dx}$

Given that $\arccos{\frac{x^{2}-1}{x^{2}+1}} = \arccos{\frac{1-x^{2}}{1+x^{2}}}$ and put $x=\tan{\theta}$

$\frac{1-\tan^{2}{\theta}}{1+\tan^{2}{\theta}}=\cos{2\theta}$ so that $y = \arccos {\cos{2\theta}}=2\theta$

$\frac{dy}{dx} = 2\frac{d}{dx}(\arctan{x})=\frac{2}{1+x^{2}}$ which is the required answer.

Question 5:

If $\arcsin{(\frac{x^{2}-y^{2}}{x^{2}+y^{2}})}=a$, where a is a parameter, then find $\frac{dy}{dx}$.

Given that $a=\arcsin{(\frac{x^{2}-y^{2}}{x^{2}+y^{2}})}$ so that $\sin{a} = \frac{x^{2}-y^{2}}{x^{2}+y^{2}}$

$(x^{2}+y^{2})\sin{a} = x^{2}-y^{2}$

Differentiating both sides w.r.t. x, we get

$2x\sin{a} + \sin{a} ,2y.\frac{dy}{dx}=2x-2y\frac{dy}{dx}$

$\frac{dy}{dx}(2y\sin{a}+2y)=2x-2x\sin{2a}$

$\frac{dy}{dx} = \frac{2x(1-\sin{a})}{1+\sin{a}}=\frac{x}{y}\times \frac{2y^{2}}{2x^{2}}=\frac{y}{x}$

Question 6:

If $y = cot^{-1}{(\sqrt{\frac{1+x}{1-x}})}$ then find $\frac{dy}{d(\arccos{x})}$.

Given that $y = cot^{-1}(\sqrt{(\frac{1+x}{1-x})})$ so that $y = \arctan{(\sqrt{(\frac{1-x}{1+x})})} = \arctan{(cot {(2\theta)})}$ where $x=\tan^{2}{\theta}$ so that $\frac{d\theta}{dx} = \frac{1}{1+x^{2}}$

and $\sec^{2}{y}.\frac{dy}{dx} = - cosec^{2}{(2\theta)}.2\frac{d\theta}{dx}$

Let $f=\arccos{x}$ so that $\frac{df}{dx} = - \frac{1}{\sqrt{1-x^{2}}}$

Now, note that $\sec^{2}{y} = cosec^{2}{2\theta}$ so we get the following simplification:

$\frac{dy}{dx} = - \frac{2}{1+x^{2}}$

Now, $\frac{dy}{df} = -\frac{\frac{dy}{dx}}{\frac{df}{dx}}= \frac{2\sqrt{1-x^{2}}}{1+x^{2}}$

Cheers,

Nalin Pithwa

## Derivatives: part 13: IITJEE Math tutorial problems for practice

Question 1:

Let $f(x)$ be a differentiable function w.r.t. x at $x=1$ and $\lim_{h \rightarrow 0} \frac{1}{h}f(1+h)=5$, then evaluate $f^{'}(1)$

Solution 1:

By definition, derivative of a function $f(x)$ is $f^{'}(x) = \lim_{t \rightarrow x}\frac{f(t)-f(x)}{t-x}$, where let us substitute $t-x=h$, $t=x+h$, as $t \rightarrow x$, then $h \rightarrow 0$

So that above expression is equal to $\lim_{h \rightarrow 0} \frac{f(x+h)-f(x)}{h}$

$f^{'}(1) = \lim_{h \rightarrow 0} \frac{f(1+h)-f(1)}{h}$ exists and can be evaluated if we know the value of the function $f(x)$ at $x=1$.

Question 2:

If $x\sqrt{1+y} + y \sqrt{1+x}=0$, then find $\frac{dy}{dx}$.

Given that $x\sqrt{1+y} + y\sqrt{1+x}=0$

Taking derivative of both sides w.r.t. x, we get the following equation:

$\sqrt{1+y} \times 1 + \frac{x}{2\sqrt{1+y}}.\frac{dy}{dx} + \sqrt{1+x}\frac{dy}{dx}+\frac{y}{2\sqrt{1+x}}.1=0$

$\sqrt{1+y}+ \frac{y}{2\sqrt{1+x}}+\frac{dy}{dx} \times (\frac{x}{2\sqrt{1+y}}+\sqrt{1+x})=0$

This further simplifies to :

$\frac{dy}{dx}. (\frac{x+2\sqrt{(1+x)(1+y)}}{2\sqrt{1+y}}) = \frac{2\sqrt{(1+x)(1+y)}+y}{2\sqrt{1+x}}$

$\frac{}{} = 2 \times \sqrt{\frac{1+y}{1+x}} \times \frac{2\sqrt{(1+x)(1+y)+y}}{x+2\sqrt{(1+x)(1+y)}}$

But, we already know that $x\sqrt{1+y}=-y\sqrt{1+x}$ so that $\sqrt{\frac{1+y}{1+x}} = - \frac{y}{x}$

$\frac{dy}{dx} = - \frac{2y}{x} \times \frac{2\sqrt{(1+x)(1+y)}+y}{x+2\sqrt{(1+x)(1+y)}}$

$\frac{dy}{dx}= -2 \times \frac{2y\sqrt{(1+x)(1+y)}+y^{2}}{x^{2}+2x\sqrt{(1+x)(1+y)}}$

$\frac{dy}{dx}=-2. \frac{2x(1+y)+y^{2}}{x^{2}-2y(1+y)}$ is the desired answer.

You can see how ugly it looks. Is there any way to simplify above? Let us give it one more shot. As follows:

Given that $x\sqrt{1+y} + y\sqrt{1+x}=0$ Hence, $x^{2}(1+y)=y^{2}(1+x)$ so that

$x^{2}-y^{2}=y^{2}x-x^{2}y$

$(x+y)(x-y) = y^{2}x-x^{2}y=xy(y-x)$. If $x \neq y$, then

$x+y= -xy$. Taking derivative of both sides w.r.t. x, we get:

$1+\frac{dy}{dx} = y(-1)+(-x)\frac{dy}{dx}$

$(1+x)\frac{dy}{dx} = -1-y$

$\frac{dy}{dx} = - \frac{1+y}{1+x}= - \frac{y^{2}}{x^{2}}$ which is such an elegant answer 🙂

Question 3:

If $x^{y}.y^{x}=c$, where c is a parameter constant, then find $\frac{dy}{dx}$ at $(e,e)$.

Solution 3:

Let $u=x^{y}$ and $v=y^{x}$.

Taking logarithm of both sides:

$\log {u} = y \log {x}$ and $\log {v} = x \log{y}$.

Consider the LHS equation:

Taking derivative of both sides w.r.t.x, we get:

$\frac{1}{u}\frac{du}{dx} = \frac{y}{x} + \frac{dy}{dx}. \log{x}$

$\frac{1}{u}. \frac{du}{dx} = \frac{y}{x} + \frac{dy}{dx} (\log{x})$

$\frac{du}{dx} = x^{y} \times (\frac{y}{x}+(\log{y}).\frac{dy}{dx})$

$\log {v} = x \log{y}$

$\frac{1}{v}\frac{dv}{dx} = \frac{x}{y}\frac{dy}{dx} + (\log{y})$

$\frac{dv}{dx} = y^{x} \times (\frac{x}{y}\frac{dy}{dx}+ \log{y})$.

Also, $x^{y}\frac{dv}{dx} + y^{x}\frac{du}{dx}=0$

$x^{y}y^{x} \times (\frac{x}{y}\frac{dy}{dx}+\log{y}) + y^{x}.x^{y}. (\frac{y}{x}+(\log{x}).\frac{dy}{dx}) =0$

$x^{y}y^{x} \times (\frac{dy}{dx}(\frac{x}{y}+\log{x})+\log{y}+\frac{y}{x})=0$ Now substitute $(x,y)=(e,e)$ and get the required answer.

$x^{y}y^{x} \times (\frac{dy}{dx}(\frac{x}{y})+\log{x}) = - x^{y}y^{x}(\frac{y}{x}+\log{y})$

$\frac{dy}{dx} (\frac{x}{y}+\log{x})=-(\frac{y}{x}+\log{y})$

Substituting $(x,y) = (e,e)$

Hence, then, $(\frac{dy}{dx})_(e,e) = - \frac{e/e + log e}{e/e + \log {e}}=-1$ is the desired answer.

Question 4:

Find $\frac{d}{dx}(\tan(\arctan{x} + cot^{-1}(x+1))$

Consider $\tan(A+B) = \frac{\tan{A}+\tan{B}}{1-\tan{A}\tan{B}}$

Subsituting $A= \arctan{x}$ and $B=cot^{-1}(x+1)$, we get the following:

$\tan{(A+B)} = \frac{tan(\arctan{x}+tan(cot^{-1}(x+1)))}{1-x. tan(cot^{-1}(x+1))} = \frac{x+\frac{1}{(x+1)}}{1-\frac{x}{x+1}}$

which in turn equals $\frac{x+\frac{1}{x+1}}{1-\frac{x}{x+1}}=\frac{x^{2}+x+1}{1}$ noting that $\arctan{(\frac{1}{x+1})}=cot^{-1}(x+1)$

Hence, the answer is $\frac{d}{dx}(x^{2}+x+1)=2x+1$

Question 5:

If $y = \arctan{\sqrt{\frac{1+\sin{x}}{1-\sin{x}}}}$, find $\frac{dy}{dx}$

Solution 5:

Given that $\tan{y} = \sqrt{\frac{1+\sin{x}}{1-\sin{x}}}$

$\tan^{2}{y} = \frac{1+\sin{x}}{1-\sin{x}}$. Taking derivative of both sides w.r.t. x,

$2(\tan{y})/ \sec^{2}{y}\frac{dy}{dx} = \frac{(1-\sin{x})(\cos{x})}{(1-\sin{x})^{2}}$

$2\tan{y}\sec^{2}{y}\frac{dy}{dx} = \frac{\cos{x}-\sin{x}\cos{x}+\cos{x}+\sin{x}\cos{x}}{(1-\sin{x})^{2}}$ which in turn equals

$\frac{2\cos{x}}{(1-\sin{x})^{2}}$

But, $\tan^{2}{y} = \frac{1+\sin{x}}{1-\sin{x}}$

so that $\sec^{2}{y}=1+\tan^{2}{y}=1+\frac{1+\sin{x}}{1-\sin{x}} = \frac{2}{1-\sin{x}}$

Hence, $2. \sqrt{\frac{1+\sin{x}}{1-\sin{x}}} \times \frac{2}{1-\sin{x}} \times \frac{dy}{dx} = \frac{2\cos{x}}{(1-\sin{x})^{2}}$

Hence, $\frac{dy}{dx} = \frac{1}{2} \times \frac{\cos{x}}{\sqrt{(1+\sin{x})(1-\sin{x})}} = \frac{\cos{x}}{2\sqrt{1-\sin^{2}{x}}} = \frac{1}{2}$

Question 6:

Find $\frac{d}{dx}cot^{-1}(\frac{1+\sqrt{1+x^{2}}}{x})$

Solution 6:

Let $y = cot^{-1} (\frac{1+\sqrt{1+x^{2}}}{x})$

Put $x = \sin{\theta}$ so that $\sqrt{1-x^{2}}=\sqrt{1-\sin^{2}{\theta}}=\cos{\theta}$

$\frac{1+\cos{\theta}}{\sin{\theta}} = \frac{2\cos^{2}(\theta/2)}{2\sin{\theta/2}\cos{\theta/2}} = cot (\theta/2)$

$cot^{-1}(cot{(\theta/2)}) = \theta/2$

$y=\theta/2$

$\frac{dy}{dx} = \frac{1}{2}\frac{d}{dx}( \arcsin {x} )=\frac{1}{2\sqrt{1-x^{2}}}$ where $|x|<1$

Question 7:

If $y = \arcsin{(\frac{2x}{1+x^{2}})}+sec^{-1}(\frac{1+x^{2}}{1-x^{2}})$. Find $\frac{dy}{dx}$.

Solution 7:

Let $x = \tan{\theta}$ so that $\frac{2x}{1+x^{2}} = \frac{2\tan{\theta}}{1+\tan^{2}{\theta}} = \sin{2\theta}$

so that $\arcsin{(\frac{2x}{1+x^{2}})} = \arcsin{\sin{2\theta}}=2\theta$

We now have $\frac{1-x^{2}}{1+x^{2}} = \frac{1-\tan^{2}{\theta}}{1+\tan^{2}{\theta}} = \cos{2\theta}$

so that $sec^{-1}{(\frac{1+x^{2}}{1-x^{2}})} = sec^{-1}(sec {(2\theta)}) =2 \theta$

so the desired answer is $4\frac{d\theta}{dx}=\frac{4}{1+x^{2}}$

Question 8:

If $y= \arcsin{(\frac{\sqrt{1+x}+\sqrt{1-x}}{2})}$ then find $\frac{dy}{dx}$

Solution 8:

Given that $\sin{y} = \frac{\sqrt{1+x}+\sqrt{1-x}}{2}$

$2\cos{y}\frac{dy}{dx} = \frac{1}{2\sqrt{1+x}} \times 1 + \frac{1}{2\sqrt{1-x}} \times (-1)$

$2\cos{y} \frac{dy}{dx} = \frac{\sqrt{1-x}-\sqrt{1+x}}{2\sqrt{1-x^{2}}}$

$2\cos{y} \frac{dy}{dx} = \frac{1-x-(1+x)}{(2\sqrt{1-x^{2}})(\sqrt{1-x}+\sqrt{1+x})} = - \frac{2x}{(2\sqrt{1-x^{2}})(\sqrt{1-x}+\sqrt{1+x})}$

$2\cos{y} \frac{dy}{dx} = - \frac{x}{(\sqrt{1-x^{2}})(2\sin{y})}$

$\frac{dy}{dx} = - \frac{x}{2\sqrt{1-x^{2}} (\sin{2y})}$ but $\sin{2y} = \frac{\sqrt{1+x}+\sqrt{1-x}}{2}$ and $\cos{y} = \frac{\sqrt{1+x}-\sqrt{1-x}}{2}$

so now we have $\sin{2y} = \frac{2}{4} (1+x-(1-x)) = x$

Hence, we get $\frac{dy}{dx} = - \frac{x}{2\sqrt{1-x^{2}}(x)}=-\frac{1}{2\sqrt{1-x^{2}}}$.

Question 9:

If $g(x) = x^{2}+2x+3f(x)$ and $f(0)=5$ and $\lim_{x \rightarrow 0} \frac{f(x)-5}{x}=4$, then evaluate $g^{'}(0)$.

Solution 9:

We have $g^{'}{(x)} = 2x+2+3f^{'}(x)$ and hence, $g^{'}{(0)}=2+3f^{'}{(0)}$

By definition of derivative, we have $f^{'}{(x)}= \lim_{t \rightarrow x} \frac{f(t)-f(x)}{t-x}$ where let us say $t-x=h$ so that $t \rightarrow x$, and $h \rightarrow 0$

$f^{'}(x) = \lim_{h \rightarrow 0} \frac{f(x+h)-f(x)}{h}$

$\lim_{h \rightarrow 0} \frac{f(h)-5}{h}=4$ and $f(0)=5$ and hence, $f^{'}(0)=4$.

Hence, $g^{'}(0)=2+3 \times 4=14$

Question 10:

If $\tan{y} = \frac{2t}{1-t^{2}}$, and $\sin{x}=\frac{2t}{1+t^{2}}$, then find $\frac{d^{2}y}{dx^{2}}$.

Let $t = \tan{\theta}$ and hence, $\frac{2t}{1-t^{2}} = \frac{2\tan{\theta}}{1-\tan^{2}{\theta}} = \tan{2\theta}$

Hence, $\tan{y} = \tan{2\theta}$

so that $\sec^{2}{y} \frac{dy}{dx} = \sec^{2}{(2\theta)}.2.\frac{d\theta}{dx}$

Let $t=\tan{\theta}$ so that $\theta = \arctan{t}$ and $\frac{d\theta}{dx}=\frac{1}{1+t^{2}} \times \frac{dt}{dx}$

Hence, we get $(\sec^{2}{y})\frac{dy}{dx} = \sec^{2}(2\theta) \times \frac{2}{(1+t^{2})}.\frac{dt}{dx}$ so that

$\sec^{2}{(2\theta)}\frac{dy}{dx} = \sec^{2}{(2\theta)} \times \frac{2}{1+t^{2}} \times \frac{dt}{dx}$

Hence, $\frac{dy}{dx} = \frac{2}{(1+t^{2})}\frac{dt}{dx}$

Hence, $\sin{x}=\frac{2t}{1+t^{2}} = \frac{2\tan{\theta}}{1+\tan^{2}{(\theta)}}=2 \frac{\sin{\theta}}{\cos{\theta}} \frac{\cos^{2}{\theta}}{1}$

Hence, $\sin{x} = \sin{2\theta}$ and hence $x=2\theta$ and so $t=\tan{\theta}$ hence, $\theta=\arctan{t}$

$x =\arctan{t}$ so that $t=\tan{x}$

$\frac{dt}{dx} = \frac{1}{1+x^{2}}$….call this A.

$\frac{dy}{dx} = \frac{2}{1+t^{2}} \frac{dt}{dx}$…call this B.

$\frac{dy}{dx} = \frac{2}{(1+t^{2})(1+x^{2})}$

$\frac{d^{2}y}{dx^{2}} = \frac{d}{dx}(\frac{2}{(1+t^{2})(1+x^{2})}) = \frac{2}{(1+t^{2})} \frac{d}{dx}(\frac{1}{1+x^{2}}) + \frac{2}{(1+x^{2})} \frac{d}{dx}(\frac{1}{(1+t^{2})})$

$\frac{d^{2}y}{dx^{2}} = \frac{2}{(1+t^{2})}. \frac{-2x}{(1+x^{2})^{2}}+\frac{2}{(1+x^{2})}.\frac{-2t}{(1+t^{2})^{2}}.\frac{dt}{dx}$

$=\frac{-4x}{(1+t^{2})(1+x^{2})^{2}} - \frac{4t}{(1+x^{2})^{2}(1+t^{2})^{2}}$

$\frac{d^{2}y}{dx^{2}} = \frac{-4x(1+t^{2})-4t}{(1+x^{2})^{2}(1+t^{2})^{2}} = \frac{-4(t+x(1+t^{2}))}{(1+x^{2})^{2}(1+t^{2})^{2}}$ which in turn equals

$\frac{-4(\tan{x}+x (1+\tan^{2}{x}))}{(1+x^{2})^{2}(1+\tan^{2}{x})^{2}}$ so where did we go wrong….quite clearly, practice alone can help us develop foresight…below is a cute proof:

$\frac{dy}{dx}= \frac{\frac{dy}{dt}}{\frac{dx}{dt}} = \frac{\frac{d}{dt}(\arctan{(\frac{2t}{1-t^{2}})})}{\frac{d}{dt}\arcsin(\frac{2t}{1+t^{2}})}$ and put $t=\tan{\theta}$

so that $\frac{dy}{dx} = \frac{2\frac{d\theta}{dt}}{2\frac{d\theta}{dt}}=1$ so we have bingo 🙂 an elegant answer

$\frac{d^{2}y}{dx^{2}}=1$

Cheers,

Nalin Pithwa.

## Derivatives: part 12:IITJEE maths tutorial problems for practice

1, $x = a(t+\frac{1}{t})$, $y=a(t-\frac{1}{t})$, then find $\frac{dy}{dx}$.

Option (A) $\frac{t^{2}-1}{t^{2}+1}$

Option (B) $\frac{t^{2}+1}{t^{2}-1}$

Option (C) $\frac{t^{2}+1}{1-t^{2}}$

Option (D) $\frac{1}{t}$

Solution 1: Given that $x=at+\frac{a}{t}$ so that $\frac{dx}{dt} = a +\frac{a(-1)}{t^{2}} = a(1-\frac{1}{t^{2}})$

and given that $y = at - \frac{a}{t}$ so that $\frac{dy}{dt} = x + \frac{a}{t^{2}} = a(1+ \frac{1}{t^{2}})$

and so we get $\frac{dy}{dx} = \frac{\frac{dy}{dt}}{\frac{dx}{dt}} = \frac{a(1 - \frac{1}{t^{2}})}{a(1+\frac{1}{t^{2}})} = \frac{t^{2}-1}{t^{2}+1}$ so that correct choice is option A.

2. If $x = a \sin{3t} + b \cos{3t}$ and $y= b \cos {t} + a \sin{t}$ then find $\frac{dy}{dx}$ when $t = \frac{\pi}{4}$

Option (A) 0

Option (B) $\frac{b-a}{3(a+b)}$

Option (C) $\frac{a-b}{3(a+b)}$

Option (D) $\frac{b-a}{b+a}$

Solution 2:

$\frac{dy}{dt} = -b \sin{t} + a\cos{t}$

$\frac{dx}{dt} = 3a \cos{3t} + -3b \sin{3t}$

$\frac{dy}{dx} = \frac{dy/dt}{dx/dt} = - \frac{-b\sin{t}+a \cos{t}}{3a \cos{3t}-3b\sin{3t}}$ which is equal to the following at $t = \frac{\pi}{4}$

$\frac{dy}{dx} = \frac{- \frac{b}{\sqrt{2}} + \frac{a}{\sqrt{2}}}{-\frac{3a}{\sqrt{2}} - \frac{3b}{\sqrt{2}}}=-\frac{1}{3} \times \frac{b-a}{a+b}$ so that the correct choice is C.

3. If $y = \frac{x \arcsin{x}}{\sqrt{1-x^{2}}} + \log{(\sqrt{1-x^{2}})}$, then find $\frac{dy}{dx}$

Option A: $\frac{\arcsin{x}}{(1-x^{2})^{\frac{3}{2}}}$

Option B: $\frac{\arcsin{x}}{\sqrt{1-x^{2}}}$

Option C:$\frac{\arcsin{x}}{1-x^{2}}$

Option D: $(1-x^{2})^{\frac{3}{2}}\arcsin{x}$

Solution 3:

Let $y = f(x) + g(x)$ where we put $f(x) = \frac{x \arcsin{x}}{\sqrt{1-x^{2}}}$ so now let $x=\sin{\theta}$

So, we get $\frac{dx}{d\theta} = \cos{\theta}$ and $1-x^{2}= \cos^{2}{\theta}$ and $\sqrt{1-x^{2}} = \cos{\theta}$

So we get $f(\theta) = \frac{\theta \times \sin{\theta}}{\cos{\theta}} = \theta \times \tan{\theta}$

So now $\frac{df}{d\theta}= \tan{\theta}+ \theta \times \sin^{2}{\theta}$

And, $g(x) = \log{\sqrt{1-x^{2}}}$

$\frac{dg}{dx} = \frac{1}{\sqrt{1-x^{2}}} \times \frac{d}{dx} (\sqrt{1-x^{2}}) = \frac{1}{2(1-x^{2})} \times (-2x)= \frac{-x}{1-x^{2}}$

Hence, we get the following:

$\frac{dy}{dx} = \frac{x}{\sqrt{1-x^{2}}} + \frac{\arcsin{x}}{\frac{1}{1-x^{2}}} + \frac{-x}{1-x^{2}}$

Question 4: Find the following: $\frac{d}{dx}(sec^{-1}{(\frac{1}{\sqrt{1-x^{2}}})} + cot^{-1}(\frac{\sqrt{1-x^{2}}}{x}))$

Option a: $\frac{2}{\sqrt{1-x^{2}}}$

Option b: $\frac{1}{\sqrt{1-x^{2}}}$

Option c: $\frac{\sqrt{1-x^{2}}}{x}$

Option d: $\sqrt{1-x^{2}}$

Solution 4:

Let $f(x)=y_{1}=sec^{-1}(\frac{1}{\sqrt{1-x^{2}}})$

Let $x=\sin{\theta}$, $1-x^{2}=\cos^{2}(\theta)$, $\sqrt{1-x^{2}}=\cos{\theta}$, and $\frac{1}{\sqrt{1-x^{2}}} = \sec{\theta}$

so $y_{1}=\theta=\arcsin{x}$

$\frac{dy_{1}}{dx} = \frac{1}{\sqrt{1-x^{2}}}$

Let $y_{2}=cot^{-1}{\frac{\sqrt{1-x^{2}}}{x}}$

Let $x=\sin{\theta}$, $\sqrt{1-x^{2}}=\cos{\theta}$ and $\frac{\sqrt{1-x^{2}}}{x}=\cot{\theta}$

Let $y_{2}= \cot^{-1}{\cot{\theta}}=\theta$

so that $\frac{dy_{2}}{dx} = \frac{d}{dx}(\arcsin{x})=\frac{1}{1-x^{2}}$

so that $\frac{dy}{dx}=\frac{2}{\sqrt{1-x^{2}}}$ so the option is a.

Question 5:

If $y=(x+\sqrt{1+x^{2}})^{n}$ then find $(x^{2}+1)(\frac{dy}{dx})^{2}$.

Solution 5:

$y=(x+\sqrt{1+x^{2}})^{n}$

$\frac{dy}{dx} = x(x+\sqrt{1+x^{2}})^{n-1}\frac{d}{dx}(x+\sqrt{1+x^{2}}) = n(x+\sqrt{1+x^{2}})^{n-1}(1+\frac{2x}{2\sqrt{1+x^{2}}})$

$(\frac{dy}{dx})^{2}.(x^{2}+1) = (x^{2}+1).n^{2}.(x+\sqrt{1+x^{2}})^{2n-2} \times (1+\frac{x}{\sqrt{1+x^{2}}})^{2}$

$=(x^{2}+1).n^{2}. (x+\sqrt{1+x^{2}})^{2n-2} \times (\frac{x+\sqrt{1+x^{2}}}{\sqrt{1+x^{2}}})^{2}$

$=(x^{2}+1).n^{2}.(x+\sqrt{1+x^{2}})^{2n}.\frac{1}{(1+x^{2})}=n^{2}y^{2}$

Question 6:

If $f(x)= \sqrt{\frac{(x+1)(x+2)}{(x+3)(x+6)}}$, then $f^{'}(0)$ is equal to

(a) 1/2 (b) 1/3 (c) 1/6 (d) 0

Solution 6:

Given that $y = \sqrt{\frac{(x+1)(x+2)}{(x+3)(x+6)}}$

Hence, we have $(x+3)(x+6) y^{2}=(x+1)(x+2)$

$(x+3)\frac{d}{dx}(y^{2}(x+6))+y^{2}(x+6) \times 1 = (x+1) \times 1 + (x+2) \times 1=2x+3$

$(x+3)(x+6)2y\frac{dy}{dx} + (x+3)y^{2} \times 1 =2x+3$

$(x+3)(x+6)2y\frac{dy}{dx} + (x+3) \times \frac{(x+1)(x+2)}{(x+3)(x+6)} = 2x+3$

$(x+3)(x+6)^{2}.2. \sqrt{\frac{(x+1)(x+2)}{(x+3)(x+6)}}+(x+1)(x+2) = (2x+3)(x+6)$

$2\sqrt{(x+1)(x+2)(x+3)(x+6)} \times (x+6)\frac{dy}{dx} + (x+1)(x+2) = (2x+3)(x+6)$

So, at x=0, on substitution we get $f^{'}(0)$.

Question 7:

If $y = \frac{1-t^{2}}{1+t^{2}}$, $x = \frac{2t}{1+t^{2}}$, then find $\frac{dy}{dx}$.

Solution 7:

Given $y= \frac{1-t^{2}}{1+t^{2}}$, let $t= \tan{\theta}$ so that $\frac{dt}{d\theta}= \sec^{2}(\theta)$

so that $y = \frac{1-t^{2}}{1+t^{2}} = \frac{1-\tan^{2}{\theta}}{1+\tan^{2}{\theta}} = \frac{\cos^{2}{\theta}-\sin^{2}{\theta}}{1} = 2 \cos^{2}{\theta}-1= \cos{2\theta}$

Now, $x = \frac{2t}{1+t^{2}}=\frac{2\tan{\theta}}{1+\tan^{2}{\theta}}$ so that $x = \sin{2\theta}$

so now $\frac{dx}{d\theta}=2 \cos{2\theta}$

$y = \cos{2\theta}$

$\frac{dy}{d\theta} = -2 \sin{2\theta}$

$\frac{dy}{dx} = - \frac{2\sin{(2\theta)}}{2(\cos{2\theta})}= - \tan{(2\theta)} = - \frac{2t}{1-t^{2}}= - \frac{x}{y}$.

Question 8:

Find $\frac{d}{dx}(\arctan{x} + \arcsin{(\frac{x}{\sqrt{1+x^{2}}})})$

Solution 8:

Let it be given that $y = \arctan{x} + \arcsin{(\frac{x}{\sqrt{1+x^{2}}})}$

Now, let us simplify this as $y=y_{1}+y_{2}$ where $y_{1} = \arctan{x}$ and $y_{2} = \arcsin{(\frac{x}{\sqrt{1+x^{2}}})}$

Now, first consider $y_{1} = \arctan{x}$. Taking derivative of both sides w.r.t. x, we get

$\frac{dy_{1}}{dx} = \frac{d}{dx}(\arctan{x}) = \frac{1}{1+x^{2}}$….A

Now, next consider $y_{2} = \arcsin{(\frac{x}{\sqrt{1+x^{2}}})}$. Takind derivative of both sides w.r.t. x, we get

$\frac{dy_{2}}{dx} = \frac{d}{dx}(\arcsin{(\frac{x}{\sqrt{1+x^{2}}})}) = \frac{1}{1- \frac{x^{2}}{1+x^{2}}} \frac{d}{dx}(\frac{x}{\sqrt{1+x^{2}}}) = \frac{1}{1+x^{2}}(\frac{1}{\sqrt{1+x^{2}}+\frac{x}{2}(1+x^{2})^{-3/2}})$….B

So that we get $\frac{dy}{dx} = \frac{dy_{1}}{dx} + \frac{dy_{2}}{dx}$using A and B.

Question 9:

If $x^{y} = y^{x}$, then find $\frac{dy}{dx}$

Solution 9:

Given that $x^{y} = y^{x}$

$y \log{x}= x \log{y}$. Taking derivative of both sides w.r.t. x, we get

$(\log{x}).\frac{dy}{dx} + \frac{y}{x} = \frac{x}{y}. \frac{dy}{dx} + (\log{y}) \times 1$

$(\log{x}- \frac{x}{y}).\frac{dy}{dx} = (\log{y}) - \frac{y}{x}$

$\frac{dy}{dx} = \frac{\frac{x(\log{y}-y)}{x}}{\frac{y\log{x}-x}{y}}= \frac{y}{x} \times \frac{x(\log{y})-y}{y(\log{x})-x}$ which is the required answer.

Question 10:

If $(x+y)^{m+n} = x^{m}y^{n}$, then find $\frac{dy}{dx}$.

Solution 10:

Given that $(x+y)^{m+n} = x^{m}y^{n}$

Taking logarithm of both sides w.r.t. any arbitrary valid base,

$(m+n) \times \log{(x+y)} = \log{(x^{m}y^{n})} = \log(x^{m}) + \log{y^{n}}$ so that $(m+n).\log{(x+y)}=m \log{x} + n \log{y}$

Taking derivative of both sides w.r.t. x, we get the following:

$\frac{m+n}{x+y}. \times (1+\frac{dy}{dx}) = \frac{m}{x} + \frac{n}{y}.\times \frac{dy}{dx}$

$\frac{m+n}{x+y} \times \frac{dy}{dx} - \frac{x}{y}. \frac{dy}{dx} = \frac{m}{x} - \frac{m+n}{x+y}$

$\frac{(m+n)y-n(x+y)}{y(x+y)}. \frac{}{} = \frac{mx+my-mx-nx}{x(x+y)}$

$\frac{my+ny-nx-ny}{} = \frac{my-nx}{x(x+y)}$

$\frac{my-nx}{y(n+y)}. \frac{dy}{dx} = \frac{my-nx}{x(x+y)}$, so that finally we get the desired answer:

$\frac{dy}{dx} = \frac{y}{x}$

More later,

Cheers,

Nalin Pithwa

## Derivatives: part 11: IITJEE maths tutorial problems for practice

Problem 1: Find $\frac{d}{dx}\arctan{\frac{4x}{4-x^{2}}}$.

Choose (a) $\frac{1}{4-x^{2}}$ (b) $\frac{1}{4+x^{2}}$ (c) $\frac{4}{4+x^{2}}$ (d) $\frac{4}{4-x^{2}}$

Solution 1:

Let $y = \arctan{\frac{4x}{4-x^{2}}}$. Hence, $\tan{y} = \frac{4x}{4-x^{2}}$. Differentiating both sides w.r.t. x, we get the following:

$\sec^{2}{y} \times \frac{dy}{dx}= \frac{d}{dx} (\frac{4x}{4-x^{2}})$

$\sec^{2}{y} \times \frac{dy}{dx} = \frac{(4-x^{2}) \times 4 - 4x \times (-2x)}{(4-x^{2})^{2}} = \frac{16+4x^{2}}{(4-x^{2})}$

But, $\sec^{2}{y}=\tan^{2}{y}+1=\frac{(x^{2}+4)^{2}}{(4-x^{2})^{2}}$

Hence, the answer is $\frac{dy}{dx}= \frac{4}{4+x^{2}}$. Option c.

Problem 2: Find $\frac{dy}{dx}$ if $\sqrt{\frac{x}{y}} + \sqrt{\frac{y}{x}}=1$

Choose (a) $\frac{-(2x+y)}{x+y}$ (b) $\frac{-(2x+y)}{x+2y}$ (c) $\frac{x+2y}{x+y}$ (d) $-\frac{2x+y}{x+2y}$

Solution 2:

The given equation is $x+y = \sqrt{xy}$. Differentiating both sides wrt x,

$1+ \frac{dy}{dx} = \sqrt{y} \times \frac{d}{dx} (\sqrt{x})+ \sqrt{x}\frac{d}{dx}(y^{\frac{1}{2}})$

$1+\frac{dy}{dx} = \frac{\sqrt{y}}{2\sqrt{x}} + \frac{\sqrt{x}}{2\sqrt{y}} \times \frac{dy}{dx}$

$(1- \frac{1}{2}\sqrt{\frac{x}{y}}) \times \frac{dy}{dx} =\frac{\sqrt{y}}{2\sqrt{x}} -1$

$\frac{dy}{dx} = \frac{\sqrt{y}-2\sqrt{x}}{2\sqrt{y}-\sqrt{x}} \times \frac{2\sqrt{y}}{2\sqrt{x}}$

$\frac{}{} = \frac{2y - 4\sqrt{xy}}{4\sqrt{xy}-2x} = \frac{2y-4(x+y)}{4(x+y)-2x} = - \frac{2x+y}{x+2y}$ is the answer. Option D.

Problem 3: If $y=\arctan{\frac{\log{(\frac{e}{x^{2}})}}{\log{(ex^{2})}}}$ then $\frac{dy}{dx}$ is

choose (a) $e$ (b) $\frac{2}{x(1+4(\log{x})^{2})}$(c) $\frac{-2}{x(1+4(\log{x})^{2})}$ (d) $\frac{2}{1+x^{2}}$

Solution 3:

Given that $y = \arctan{(\frac{\log(\frac{e}{x^{2}})}{\log(ex^{2})})}$ so that we have

$\tan{y} = \frac{\log{\frac{e}{x^{2}}}}{\log{ex^{2}}}$ so now differentiating both sides w.r.t. x,

$\sec^{2}{y}\frac{dy}{dx} = \frac{\frac{\log{(ex^{2})}}{\frac{e}{x^{2}}} \frac{d}{dx}(\frac{e}{x^{2}}) - \log{(\frac{e}{x^{2}})} \times \frac{1}{ex^{2}} \times \frac{d}{dx}(ex^{2})}{(\log{(ex^{2})})}$

$\sec^{2}{y}\frac{dy}{dx} = \frac{-\frac{2}{x}(\log{(ex^{2})})- \frac{2}{x}\log{(\frac{e}{x^{2}})}}{(\log{(ex^{2})})^{2}}$

$\sec^{y}(\frac{dy}{dx}) = \frac{-\frac{2}{x}(\log{(ex^{2}) \times (\frac{e}{x^{2}})})}{(\log{ex^{2}})^{2}}$

$\sec^{y}\frac{dy}{dx} = \frac{-\frac{4}{x}}{(\log{(ex^{2})^{2}})} = \frac{-4}{x(\log{(ex^{2})})^{2}}$

Now, we also know that$\sec^{2}{y} = 1 + \tan^{2}{y} = \frac{(\log{(\frac{e}{x^{2}})})^{2}}{(\log{(ex^{2})})^{2}} + 1 = \frac{(\log(\frac{e}{x^{2}}))^{2}+(\log{(ex^{2})})^{2}}{(\log{(ex^{2})})^{2}}$

But, note that by laws of logarithms, on simplification, we get

$\log{(\frac{e}{x^{2}})} = 1 - 2\log{x}$ and $\log{(ex^{2})} = 1 + 2 \log{x}$ so that on squaring, we get

$(\log{(e/x^{2})})^{2} = 1-4\log{x} + 4 (\log{x})^{2}$

$(\log{(ex^{2})})^{2}=1+4\log{x} + 4 (\log{x})^{2}$ so that now we get

$(\log{(\frac{e}{x^{2}})})^{2} + (\log{(ex^{2})})^{2} = 2 + 8 (\log{x})^{2}$, which all put together simplifies to

$\frac{dy}{dx} = \frac{1}{\sec^{2}{y}} \times \frac{-(\frac{4}{x})}{(\log{(ex^{2})})^{2}}$

$\frac{dy}{dx} = - \frac{(\frac{2}{x})}{1+4(\log{x})^{2}}$ so that the answer is option C.

Problem 4: Find $\frac{d}{dx}(\arcsin{(3x-4x^{3})}+\arccos{(2x(\sqrt{(1-x^{2})}))})$

Choose option (a) $\frac{1}{\sqrt{1-x^{2}}}$ (b) $\frac{-1}{\sqrt{1-x^{2}}}$ (c) $\frac{5}{\sqrt{1-x^{2}}}$ (d) $\frac{-2}{\sqrt{1-x^{2}}}$

Solution 4:

Let us consider the first differential. Let us substitute $x = \sin{\theta}$. Hence,

$3x-4x^{3}=3\sin{\theta} - 4\sin^{3}{\theta}= \sin{3\theta}$ and so we $\arcsin{3x-4x^{3}} = \arcsin{\sin{3\theta}} = 3 \theta$, and so also, we get $\arccos{2x\sqrt{1-x^{2}}}=\arccos{2\sin{\theta}\cos{\theta}} = \arccos{\cos{2\theta}}=2\theta$ so we get

required derivative

$\frac{dy}{dx} = \frac{d}{dx}(3\theta) + \frac{d}{dx}(2\theta) = \frac{d}{dx}(5\theta) = 5 \frac{d\theta}{dx} = 5 \frac{d}{dx}(\arcsin{x})= 5 \frac{1}{\sqrt{1-x^{2}}}$. Answer is option C.

Problem 5: Find $\frac{d}{dx}(x-a)(x-b)(x-c)\ldots (x-z)$

Choose option (a) zero (b) 26 (c) 26! (d) does not exist

Solution 5: the expression also includes a term $0 = (x-x)$ so that the final answer is zero only.

Problem 6: Find $\frac{d}{dx}(x^{x})^{x}$.

Solution 6: Let $y= (x^{x})^{x}$

so $\log{y} = \log{(x^{x})^{x}}$

so $\log{y} = x^{2} \times \log{x}$ so that differentiating both sides w.r.t. x, we get

we get $\frac{1}{y} \times \frac{dy}{dx} = \frac{x^{2}}{x} + \log{x} \times 2x$

we get $\frac{1}{y} \times \frac{dy}{dx} = x + 2x \log{x} = x(1+2\log{x})$

we get $\frac{dy}{dx} = yx (1+2 \log{x}) = (x^{x})^{x} \times x \times (1+2\log{x})$

so the answer is option B.

Choose option (a): $x.x^{x}(1+2\log{x})$ (b) $x^{x^{2}+1} \times (1+2\log{x})$ (c) ${x^{{x}^{2}}}(1+\log{x})$ (d) none of these

Problem 7:

Find $\frac{d}{dx}(e^{x^{x}})$

Choose option (a) $e^{x^{x}}.x^{x}.(1+\log{x})$ (b) $e^{x^{x}}. x^{x}.\log{(\frac{x}{e})}$ (c) $e^{x^{x}}.x^{x}$ (d) $e^{x^{x}}. (\log{(e^{x})})$

Solution 7: Let $y = (e^{(x^{x})})$ so that taking logarithm of both sides

$\log{y} = \log{(e^{(x^{x})})}$ so that $\log{y} = x^{x} \log{e} = x^{x}$

$\log {(\log{y})}= x \times (\log{x})$. Differentiating both sides w.r.t.x we get:

$\frac{1}{\log{y}} \times \frac{d}{dx} \times (\log{y})= \frac{x}{x} + \log{x}$ so that we get now

$latex\frac{1}{y(\log{y})} \times \frac{dy}{dx} = 1 + \log{x}$

$\frac{dy}{dx} = e^{x^{x}} \times x^{x} \times (1+\log{x})$ so we get option a as the answer.

Problem 8:

Find $\frac{d}{dx}(x^{x^{x}})$

Choose option (a): $x^{x^{x}} \times (1+\log{x})$ (b) $x^{x^{x}} \times (x^{x}\log{x})(1+\log{x}+\frac{1}{x})$ (c) $x^{x^{x}} \times (x^{x}\log{x}) \times (1+\log{x}+\frac{1}{x\log{x}})$ (d) none of these.

Solution 8:

let $y=x^{x^{x}}$ taking logarithm of both sides we get

$\log{y} = x^{x} \times \log{x}$ and now differentiating both sides w.r.t.x, we get

$\frac{1}{y} \times \frac{dy}{dx} = \frac{x^{x}}{x} + (\log{x}) \times \frac{d}{dx} (x^{x})$ and now let $t=x^{x}$ and again take logarithm of both sides so that we get (this is quite a classic example…worth memorizing and applying wherever it arises):

$\log{t}= x\log{x}$

$\frac{1}{t} \frac{dt}{dx} = \frac{x}{x} + \log{x}=1+\log{x}$

$\frac{dt}{dx} = x^{x}(1+\log{x})$

$\frac{dy}{dx} \times \frac{1}{y} = x^{x-1} + (\log{x}).x^{x}.(1+\log{x})$

$\frac{dy}{dx} = x^{x}(x^{x-1}+x^{x} \times \log{x} \times (1+\log{x}))$

$\frac{dy}{dx} = x^{x^{x}} (x^{x} \times (\log{x})) \times (1+ \log{x}+ \frac{1}{x \log{x}})$

Problem 9:

Find $\frac{d}{dx}(x+a)(x^{2}+a^{2})(x^{4}+a^{4})(x^{8}+a^{8})$.

Choose option (a): $\frac{15x^{16}-16x^{15}a+a^{16}}{(x-a)^{2}}$ (b) $\frac{x^{16}-a^{16}}{x-a}$ (c) $\frac{x^{16}-x^{15}a+a^{16}}{(x-a)^{2}}$ (d) none of these

Solution 9:

Given that $y = (x+a)(x^{2}+a^{2})(x^{4}+a^{4})(x^{8}+a^{8})$

Remark: Simply multpilying out thinking the symmetry will simplify itself is going to lead to a mess…because there will be no cancellation of terms …:-) The way out is a simple algebra observation…this is why we should never ever forget the fundamentals of our foundation math:-)

note that the above can be re written as follows:

$y = \frac{(x^{2}-a^{2})}{(x-a)} \times \frac{(x^{4}-a^{4})}{(x^{2}-a^{2})} \times \frac{x^{8}-a^{8}}{(x^{4}-a^{4})} \times \frac{(x^{16}-a^{16})}{(x^{8}+a^{8})}$

Now, we are happy like little children because many terms cancel out 🙂 hahaha…lol 🙂

$y = \frac{(x^{16}-a^{16})}{(x-a)}$ and now differentiating both sides w.r.t.x we get

$\frac{dy}{dx} = \frac{(x-a)(16x^{15})- (x^{16}-a^{16})(1)}{(x-a)^{2}}$

$\frac{dy}{dx} = \frac{15x^{16}-16x^{15}a+a^{16}}{(x-a)^{2}}$

Problem 10:

If $x= \theta {\cos{\theta}}+\sin{\theta}$ and $y = \cos{\theta}-\theta \times \sin{\theta}$ then find the value of $\frac{dy}{dx}$ at$\theta = \frac{\pi}{2}$

Choose option (a): $-\frac{\pi}{2}$ (b) $\frac{2}{\pi}$ (c) $\frac{\pi}{4}$ (d) $\frac{4}{\pi}$

Solution 10:

$\frac{dx}{d\theta} = \cos{\theta} - \theta \times \sin{\theta} + \cos{\theta}$

$\frac{dy}{d\theta} = -\sin{\theta} - (\sin{\theta} + \theta \times \cos{\theta})$

$\frac{dy}{d\theta} = \theta \times \cos{\theta} - 2\sin{\theta}$

$\frac{dx}{d\theta} = 2 \cos{\theta} - \theta \times \sin{\theta}$

$\frac{dy}{dx} = \frac{\frac{dy}{d\theta}}{\frac{dy}{d\theta}} = \frac{\theta \times \cos{\theta}-2\sin{\theta}}{2\cos{\theta}-\theta \times \sin{\theta}} = \frac{-2}{-\pi/2}=\frac{4}{\pi}$

Regards,

Nalin Pithwa.

## A problem of log, GP and HP…

Question: If $a^{x}=b^{y}=c^{z}$ and $b^{2}=ac$, pyrove that: $y = \frac{2xz}{x+z}$

Solution: This is same as proving: y is Harmonic Mean (HM) of x and z;

That is, to prove that $y=\frac{2xz}{x+z}$ is the same as the proof for : $\frac{1}{y} - \frac{1}{x} = \frac{1}{z} - \frac{1}{y}$

Now, it is given that $a^{x} = b^{y} = c^{z}$ —– I

and $b^{2}=ac$ —– II
Let $a^{x} = b^{y}=c^{z}=N$ say. By definition of logarithm,

$x = \log_{a}{N}$; $y=\log_{b}{N}$; $z=\log_{c}{N}$

$\frac{1}{x} = \frac{1}{\log_{a}{N}}$; $\frac{1}{y} = \frac{1}{\log_{b}{N}}$; $\frac{1}{z} = \frac{1}{\log_{a}{N}}$.

Now let us see what happens to the following two algebraic entities, namely, $\frac{1}{y} - \frac{1}{x}$ and $\frac{1}{z} - \frac{1}{y}$;

Now, $\frac{1}{y} - \frac{1}{x} = \frac{1}{\log_{b}{N}} - \frac{1}{\log_{a}{N}} = \frac{\log_{b}{b}}{\log_{b}{N}} - \frac{\log_{a}{a}}{\log_{a}{N}} = \log_{N}{b} - \log_{N}{a} = \log_{N}{(\frac{b}{a})}$…call this III

Now, $\frac{1}{z} - \frac{1}{y} = \frac{1}{\log_{c}{N}} - \frac{1}{\log_{b}{N}} = \frac{\log_{c}{c}}{\log_{c}{}N} -\frac{\log_{b}{b}}{\log_{b}{N}}= \log_{N}{c}-\log_{N}{b}$

Hence, $\frac{1}{z} - \frac{1}{y}=\log_{N}{c/b}$….equation IV

but it is also given that $b^{2}=ac$…see equation II

Hence, $\frac{b}{a} = \frac{c}{b}$

Take log of above both sides w.r.t. base N:

So, above is equivalent to $\log_{N}{b/a} = \log_{N}{c/b}$

But now see relations III and IV:

Hence, $\frac{1}{y} -\frac{1}{x} = \frac{1}{x} - \frac{1}{y}$

Hence, $\frac{2}{y} = \frac{1}{x} + \frac{1}{z} = \frac{x+z}{xz}$

Hence, $y= \frac{2xz}{x+z}$ as desired.

Regards,

Nalin Pithwa

## Express a given integral number in any scale (radix)

Several scales (radix) have been used by mathematicians. Binary (2), Ternary (3), Quaternary (4), Quinary (5), Senary (6), Septenary (7), Octenary(8), Nonary (9), Denary (10/Decimal), Undenary(11), Duodenary (12) and of course, hexadecimal (16). Note that in any scale the base radix is “10”. Thus, “10” stands for 2 in “binary”, “ten” in “decimal”, 8 for “octal” radix respectively, etc.

Let the digits used in a proposed scale(radix r) be $a_{0}, a_{1}, a_{2}, \ldots, a_{n}$. Let us express an integer in this scale. Let $a_{0}$ be unit’s digits. Analagous to the place value system (in decimal):

$N=a_{0} + a_{1} \times r^{1} + a_{2} \times r^{2} + \ldots + a_{n} \times r^{n}$

Now, let us say we want to express this number N in terms of these digits $a_{i}$s.

Dividing N by $r$, we get the unit’s digit $a_{0}$ as the remainder; and the quotient is:

$a_{1} + a_{2} \times r^{1} + a_{3} \times r^{2} + \ldots + a_{n} \times r^{n-1}$.

Dividing the above quotient by r, we get $a_{1}$ as the remainder and the quotient as:

$a_{2} \times r^{1} + a_{3} \times r^{2} + a_{4} \times r^{3} + \ldots + a_{n} \times r^{n-2}$, and so on.

Example: Express the denary number 5213 in the scale of seven.

Solution: $(5213)_{10} \div 7$ gives 5 as remainder and $(744)_{10}$ as quotient.

$(744)_{10} \div 7$ gives 2 as remainder and $(106)_{10}$ as remainder.

Continuing this way, we are able to express:

$(5213)_{10} = 2 \times 7^{4} + 1 \times 7^{3} + 1 \times 7^{2} + 2 \times 7 +5$. That is $(21125)_{7}$. You can check the equivalence by converting both to decimal values.

Cheers,

Nalin Pithwa.