## 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.

## Derivatives: Part 10: IITJEE maths tutorial problems for practice

Problem 1: If $x=3\cos{\theta}-\cos^{3}{\theta}$, and $y=3\sin{\theta}-\sin^{3}{\theta}$, then $\frac{dy}{dx}$ is equal to:

(a) $-\cot^{3}{\theta}$ (b) $-\tan^{3}{\theta}$ (c) $\cot^{3}{\theta}$ (d) $\tan^{3}{\theta}$

Problem 2: If $x = \tan{\theta} + \cot{\theta}$, and $y=2 \log{(\cot{\theta})}$, then $\frac{dy}{dx}$ is equal to:

(a) $\tan{(2\theta)}$ (b) $\cot{(2\theta)}$ (c) $\tan{\theta}$ (d) $\sec^{2}{2\theta}$

Problem 3: $\frac{d}{dx}\log{\sqrt{\frac{1-\cos{x}}{1+\cos{x}}}}$ is equal to:

(a) $\tan{\frac{x}{2}}$ (b) $\sin{x}$ (c) cosec(x) (d) $\tan{x}$

Problem 4: $y=2^{2(\log_{2}{(x+2)}-\log_{2}{(x+1)})}$, then $\frac{dy}{dx}$ is:

(a) $\frac{-2(x+2)}{(x+1)^{3}}$ (b) $\frac{4(x+2)}{(x+1)^{3}}$ (c) $\frac{2(x+2)}{(x+1)^{3}}$ (d) $\frac{-6(x+2)}{(x+1)^{3}}$

Problem 5: $\frac{d}{dx}(\arctan{\sqrt{\frac{e^{x}-1}{e^{x}+1}}})$ is equal to:

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

Problem 6: $y=e^{m\arcsin{x}}$ then $(1-x^{2}) (y^{'})^{2}$ is equal to :

(a) $y^{2}$ (b) $m^{2}(1-y^{2})$ (c) $-m^{2}y^{2}$ (d) $m^{2}y^{2}$

Problem 7: If $y = \sin(m \arcsin{x})$ then $(1-x^{2})(\frac{dy}{dx})^{2}$ is

(a) $m^{2}y^{2}$ (b) $m^{2}(1-y^{2})$ (c) $-m^{2}y^{2}$ (d) $m^{2}(1+y^{2})$

Problem 8: $\frac{d}{dx}(\cos{\arctan{x}})$ is:

(a) $\frac{1}{2\sqrt{1+x^{2}}}$ (b) $\frac{-x}{(1+x^{2})^{\frac{3}{2}}}$

(c) $\frac{-x}{\sqrt{1+x^{2}}}$ (d) $\frac{2x}{\sqrt{1+x^{2}}}$

Problem 9: If $y = \arcsin{\frac{a\cos{x}+b\sin{x}}{\sqrt{a^{2}+b^{2}}}}$ then $\frac{d^{2}y}{dx^{2}}$ is:

(a) $-1$ (b) 0 (c) 1 (d) $\arctan{\frac{b}{a}}$

Problem 10: $\arctan{\frac{4x}{4-x^{2}}}$ is:

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

Regards,

Nalin Pithwa.

## Derivatives: Part 9: IITJEE maths tutorial problems practice

Problem 1: $\frac{d}{dx}((\frac{1}{b}\arctan{\frac{x}{b}})-\frac{1}{a}\arctan{(\frac{x}{a})})$ is equal to:

(a) $\frac{1}{(x^{2}+a^{2})(x^{2}+b^{2})}$ (b) $\frac{a^{2}-b^{2}}{(x^{2}+a^{2})(x^{2}+b^{2})}$

(c) $\frac{x^{2}+a^{2}}{x^{2}+b^{2}}$ (d) $\frac{2x^{2}}{(x^{2}+a^{2})(x^{2}+b^{2})}$

Problem 2: $\frac{d}{dx}(\frac{x}{2} + \frac{1}{2}\log{(\sin{x}+\cos{x})})$ is equal to:

(a) $\frac{\tan{x}}{1+\tan{x}}$ (b) $\frac{1}{1+\cot{x}}$ (c) $\frac{1-\tan{x}}{1+\tan{x}}$ (d) $\frac{1}{1+\tan{x}}$

Problem 3: If $y=\sqrt{\frac{cosec{x}-\cot{x}}{cosec{x}+\cot{x}}}$ where $0, then $\frac{dy}{dx}$ is given by :

(a) $cosec{x}(cosec{x}-\cot{x})$ (b) $cosec{x}(\cot{x}-cosec{x})$ (c) $cosec{x}(\cot{x}-cosec{x})$ (d) $\cot{x}(cosec{x}-\cot{x})$

Problem 4: $\frac{d}{dx}\log {|\sec{(x-\frac{\pi}{4})}+\tan{(x-\frac{\pi}{4})}|}$ is equal to:

(a) $\frac{\sqrt{2}}{\sin{x}-\cos{x}}$ (b) $\frac{\sin{x}}{\sin{x}+\cos{x}}$ (c) $\frac{\sqrt{2}}{\sin{x}+\cos{x}}$ (d) $\frac{1}{\sin{x}+\cos{x}}$

Problem 5:

If $r=a(1+\cos{\theta})$, and $\tan{\phi}=r\frac{d\theta}{dr}$, then $\phi$ is equal to:

(a) $\frac{-2}{\theta}$ (b) $\frac{\pi}{2} + \frac{\theta}{2}$ (c) $-\frac{\theta}{2}$ (d) $\frac{\pi}{2} - \frac{\theta}{2}$

Problem 6: $\frac{d}{dx}\log{(\sqrt{x+ \sqrt{x^{2}+a^{2}}})}$ is equal to:

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

Problem 7: $\frac{d}{dx}(\log{(1+\sin{(2x)})} + 2 \log{\sec{(\frac{\pi}{4}-x)}})$ is equal to

(a) 0 (b) $\log{2}$ (c) $\frac{4(\cos{x}-\tan{x})}{\sin{x}+\cos{x}}$ (d) $\frac{2\cos{(2x)}}{1+\sin{(2x)}} + \tan{(\frac{\pi}{4}-x)}$

Problem 8: If $x^{2}+xy+y^{2}=1$, then $\frac{dy}{dx}$ is equal to:

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

Problem 9: $\frac{d}{dx}(\arcsin{(\sqrt{\frac{1-x}{2}})})$ is equal to:

(a) $\frac{1}{\sqrt{1-x^{2}}}$ (b) $\frac{-1}{\sqrt{1-x^{2}}}$ (c) $\frac{1}{2\sqrt{1-x^{2}}}$ 9d) $\frac{-1}{2\sqrt{1-x^{2}}}$

Problem 10: If $y = \arctan{(\frac{3a^{2}x-x^{3}}{x^{3}-3ax^{2}})}$ then $\frac{dy}{dx}$ is equal to:

(a) $\frac{3}{a}$ (b) $\frac{1}{a}$ (c) $\frac{3x}{a}$ (d) $\frac{3a}{x^{2}+a^{2}}$

Cheers,

Nalin Pithwa