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Derivatives: part 11: IITJEE maths tutorial problems for practice

March 11, 2021 – 2:25 am

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):

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

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

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

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

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

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

The answer is option C.

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

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

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

The answer is option A.

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:

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

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

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

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

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

Answer is option D.

Regards,

Nalin Pithwa.

By Nalin Pithwa | Also posted in calculus | Comments (0)

A problem of log, GP and HP…

January 27, 2021 – 3:48 am

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

By Nalin Pithwa | Also posted in algebra, IITJEE Foundation mathematics, Pre-RMO | Comments (0)

Derivatives: Part 10: IITJEE maths tutorial problems for practice

January 18, 2021 – 5:18 am

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

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.

By Nalin Pithwa | Also posted in calculus, KVPY | Comments (0)

Derivatives: Part 9: IITJEE maths tutorial problems practice

January 17, 2021 – 11:24 pm

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

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 $0<x<\frac{\pi}{2}$ , 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

By Nalin Pithwa | Also posted in calculus | Comments (0)

Derivatives: part 8: IITJEE mains tutorial problems practice

January 17, 2021 – 3:42 pm

Problem 1: If $y=b(\arctan{(\frac{x}{y})})+ \arctan{(\frac{y}{x})}$ , then $\frac{dy}{dx}$ is equal to:

(a) $\frac{x}{2}$ (b) -1 (c) 0 (d) b

Problem 2: If $r=a(1+\cos{\theta})$ , then $\sqrt{r^{2}+(\frac{dr}{d\theta})^{2}}$ is:

(a) $2a\cos{\theta}$ (b) $2a \sin{(\frac{\theta}{2})}$ (c) $2a \cos{(\frac{\theta}{2})}$ (d) $2a \sin{\theta}$

Problem 3: $\frac{d}{dx}\arctan{\log_{10}{x}}$ is equal to:

(a) $\frac{1}{1 + (\log_{10}{x})^{2}}$ (b) $\frac{1}{x \log_{10}{(1+ (\log_{10}{x})^{2})}}$ (c) $\frac{1}{x(1+(\log_{10}{x})^{2})}$ (d) $\frac{1}{10 \log{x}(1+(\log_{10}{x})^{2})}$

Problem 4: If $\sin^{2}(mx) + \cos^{2}(ny)=a^{2}$ , then $\frac{dy}{dx}$ is equal to:

(a) $\frac{m \sin{(2mx)}}{n \sin{(2ny)}}$ (b) $\frac{n\sin{(2mx)}}{m\sin{(2ny)}}$ (c) $\frac{n\sin{(2ny)}}{m\sin{(2mx)}}$ (d) $\frac{-m\sin{(2mx)}}{n\sin{(2ny)}}$

Problem 5: $\frac{d}{dx}(\frac{\tan{x}-\cot{x}}{\tan{x}+\cot{x}})$ is equal to:

(a) $2\sin{(2x)}$ (b) $\sin{(2x)}$ (c) $-2 \sin{(2x)}$ (d) $2\cos{(2x)}$

Problem 6: If $y=\log_{5}{(\log_{5}{x})}$ then the value of $\frac{dy}{dx}$ is

(a) $\frac{1}{x \log_{5}{x}}$ (b) $\frac{1}{x \log_{5}{x}. (\log{5})^{2}}$ (c) $\frac{1}{\log{5}.x\log{x}}$ (d) $\frac{1}{x(\log_{5}{x})^{2}}$

Problem 7: $\frac{d}{dx}(ax+b)^{cx+d}$ is equal to:

(a) $(ax+b)^{cx+d}(\frac{cx+d}{ax+b} + \log{(ax+b)})$ (b) $(ax+b)^{cx+d}(\frac{cx+d}{ax+b} + c \log{(ax+b)})$ (c) $a(ax+b)^{cx+d}$ (d) none

Problem 8: $\frac{d}{dx}(\log{(\frac{\sin{(x-b)}}{\sin{(x-a)}})})$ is equal to:

(a) $\frac{\cos{(a-b)}}{\sin{(x-a)}\sin{(x-b)}}$ (b) $\frac{\sin{(b-a)}}{\sin{(x-a)}\sin{(x-b)}}$

Problem 9: If $y = \sqrt{\frac{\sec{x}+\tan{x}}{\sec{x}-\tan{x}}}$ and $0<x<\frac{\pi}{2}$ , then $\frac{dy}{dx}$ is :

(a) $\sec{x}(\sec{x}-\tan{x})$ (b) $\sec{x}(\sec{x}+\tan{x})$ (c) $\tan{x}(\sec{x}+\tan{x})$ (d) $\tan{x}(\sec{x}-\tan{x})$

Problem 10: $\frac{d}{dx}e^{ax}(a\sin{(bx)}-b\cos{(bx)})$ is equal to:

(a) $e^{ax}(\sin{(bx)})$ (b) $(a^{2}+b^{2})e^{ax}\sin{(bx)}$ (c) $e^{ax}\cos{(bx)}$ (d) $(a^{2}+b^{2})e^{ax}\cos{(bx)}$

Cheers,

Nalin Pithwa.

By Nalin Pithwa | Also posted in calculus, KVPY | Comments (0)

Derivatives: part 7: IITJEE tutorial problems practice

January 7, 2021 – 8:39 pm

Problem 1: Differential coefficient of $\sec{\arctan{x}}$ is

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

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

(a) 2 (b) -2 (c) 1 (d) -1

Problem 3: If $y = \arcsin{\sqrt{x-ax}-\sqrt{a-ax}}$ , then $\frac{dy}{dx}$ is equal to:

(a) $\frac{1}{2\sqrt{x}\sqrt{1-x}}$ (b) $\sin{(\sqrt{x})} \times \sin{(\sqrt{a})}$

Problem 4: For the differentiable function f, the value of : $\lim_{h \rightarrow 0} \frac{(f(x+h))^{2}-(f(x))^{2}}{2h}$ is equal to:

(a) $(f^{'}(x))^{2}$ (b) $\frac{1}{2}(f(x))^{2}$ (c) $f(x)f^{'}(x)$ (d) zero

Problem 5: The derivative of $\arctan{\frac{\sqrt{1+x^{2}}-1}{x}}$ w.r.t. $\arctan{(\frac{2x\sqrt{1-x^{2}}}{1-2x^{2}})}$ at $x=0$ is :

(a) $\frac{1}{8}$ (b) $\frac{1}{4}$ (c) $\frac{1}{2}$ (d) 1

Problem 6: If $x = e^{y+e^{y+e^{y+e^{y+ \ldots}}}}$ then $\frac{dy}{dx}$ is

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

Problem 7: Consider the following statements:

(1) $(\frac{f}{g})^{'} = \frac{f^{'}}{g^{'}}$ (2) $\frac{(fg)^{'}}{fg} = \frac{f^{'}}{f} + \frac{g^{'}}{g}$

(3) $\frac{(f+g)^{'}}{f+g} = \frac{f^{'}}{f} + \frac{g^{'}}{g}$ (4) $\frac{(f/g)^{'}}{f/g} = \frac{f^{'}}{f} + \frac{g^{'}}{g}$

Which of the following statements are true?

(a) 1 and 2 (b) 2 and 3 (c) 2 and 4 (d) 3 and 4

Problem 8: If $y=e^{x+3\log{x}}$ then $\frac{dy}{dx} =$

(a) $e^{x+3\log{x}}$ (b) $e^{x}.x^{2}(x+3)$ (c) $e^{x}. e^{3\log{x}}$ (d) $3x^{2}e^{x}$

Problem 9: If $y=\sin^{2}(x \deg)$ , then find the value of $\frac{dy}{dx}$ is:

(a) $\frac{\pi}{360}\sin{(2 x \deg)}$ (b) $\frac{\pi}{2}\sin{(2x\deg)}$ (c) $180 \sin {(2x\deg)}$ (d) $\frac{\pi}{180}\sin{(2x\deg)}$

Problem 10: If $y=\log_{a}{x} + \log_{x}{a} + \log_{x}{x}+ \log_{a}{a}$ then the value of $\frac{dy}{dx}$ is:

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

Cheers,

Nalin Pithwa.

By Nalin Pithwa | Also posted in calculus | Comments (0)

Derivatives: part 6: IITJEE tutorial practice problems

January 6, 2021 – 1:41 am

Problem 1:

If $\sec {(\frac{x+y}{x-y})}=a$ , then $\frac{dy}{dx}$ is (i) $\frac{x}{y}$ (ii) $\frac{y}{x}$ (iii) y (iv) $x$

Problem 2:

If $f(x) = x+ 2$ , when $-1<x<1$ ;

$f(x)=5$ , when $x=3$ ;

$f(x) = 8-x$ , when $x>3$ ; then, at $x=3$ , the value of $f^{'}(x)$ is

(a) 1 (b) -1 (c) 0 (d) does not exist.

Problem 3:

If $y = x \tan{y}$ , then $\frac{dy}{dx}$ is equal to

(i) $\frac{\tan{y}}{x-x^{2}-y^{2}}$ (ii) $\frac{\tan{y}}{y-x}$

(iii) $\frac{y}{x-x^{2}-y^{2}}$ (iv) $\frac{\tan{x}}{x-y^{2}}$

Problem 4:

If g is the inverse function of f and $f^{'}(x) = \frac{1}{1+x^{n}}$ , then $g^{'}(x)$ is equal to

(i) $1 + (g(x))^{n}$ (ii) $1+g(x)$ (iii) $1-g(x)$ (iv) $1-(g(x))^{n}$

Problem 5:

If $f(x) = \log_{x^{2}}(\log{x})$ then $f(x)$ at $x=c$ is :

(i) 0 (ii) 1 (iii) $\frac{1}{e}$ (iv) $\frac{1}{2e}$

Problem 6:

If $y = (\sin{x})^{\tan{x}}$ then $\frac{dy}{dx}$ is equal to :

(i) $(\sin{x})^{\tan{x}}(1+ \sec^{2}{x} \log{\sin{x}})$

(ii) $\tan{x}. (\sin{x})^{\tan{x}-1} \times \cos{x}$

(iii) $(\sin{x})^{\tan{x}}\sec^{2}{x} \times \log{\sin{x}}$

(iv) $\tan{x} (\sin{x})^{\tan{x}-1}$

Problem 7:

If $y = \sqrt{\sin{x}+y}$ , then $\frac{dy}{dx}$ equals:

(i) $\frac{\sin{x}}{2y-1}$ (ii) $\frac{\sin{x}}{1-2y}$ (iii) $\frac{\cos{x}}{1-2y}$

(iv) $\frac{\cos{x}}{2y-1}$

Problem 8:

If $x = \sqrt{\frac{1-t^{2}}{1+t^{2}}}$ and $y = \sqrt{\frac{\sqrt{1+t^{2}}-sqrt{1-t^{2}}}{\sqrt{1+t^{2}}+\sqrt{1-t^{2}}}}$

then the value of $\frac{d^{2}y}{dx^{2}}$ at $t=0$ is given by:

(a) 0 (b) 1/2 (c) 1 (d) -1

Problem 9:

If $x = a \cos^{3}{\theta}$ , $y = a \sin^{3}{\theta}$ , then $\sqrt{1 + (\frac{dy}{dx})^{2}}$ is equal to:

(i) $\sec^{2}{\theta}$ (ii) $\tan^{2}{\theta}$ (iii) $\sec{\theta}$ (iv) $|\sec{\theta}|$

Problem 10:

If $y = \arcsin{\sqrt{1-x}} + \arccos{\sqrt{x}}$ , then $\frac{dy}{dx}$ equals:

(a) $\frac{1}{\sqrt{x(1-x)}}$ (b) $\frac{1}{x(1+x)}$ (c) $\frac{-1}{\sqrt{x(1-x)}}$ (d) none

Regards,

Nalin Pithwa

By Nalin Pithwa | Also posted in calculus | Comments (0)

Derivatives: part 5: IITJEE maths tutorial problems for practice

November 14, 2020 – 5:09 pm

Problem 1:

The derivative of $arcsec (\frac{1}{1-2x^{2}})$ w.r.t. $\sqrt{1-x^{2}}$ at $x=\frac{1}{2}$ is

(a) 2 (b) -4 (c) 1 (d) -2

Problem 2:

If $y = \sin{\sin{x}}$ and $\frac{d^{2}y}{dx^{2}} + \frac{dy}{dx} \tan{x} + f(x)=0$ , then $f(x) =$

(a) $\sin^{2}{x} \sin{(\cos{x})}$ (b) $\cos^{2}{x}\sin{\cos{x}}$ (c) $\sin^{2}{x} \cos{\sin{x}}$ (d) $\cos^{2}{x} \sin{\sin{x}}$

Problem 3:

If $f(x) = \log_{a}{\log_{a}{x}}$ , then $f^{'}(x)$ is

(a) $\frac{\log_{a}{e}}{x \log_{e}{x}}$ (b) $\frac{\log_{e}{a}}{x}$ (c) $\frac{\log_{e}{a}}{x\log_{a}{x}}$ (d) $\frac{x}{\log_{e}{a}}$

Problem 4:

If $y=\log {\tan{\frac{x}{2}}} + \arcsin{\cos{x}}$ , then $\frac{dy}{dx}$ is

(a) $cosec (x) -1$ (b) $cosec (x) +1$ (c) $cosec (x)$ (d) x

Problem 5:

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

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

Problem 6:

Let f, g, h and k be differentiable in $(a,b)$ , if F is defined as $F(x) = \left | \begin{array}{cc} f(x) & g(x) \\ h(x) & k(x) \end{array} \right |$ for all a, b, then $F^{'}$ is given by:

(i) $\left | \begin{array}{cc} f & g \\ h & k \end{array} \right| + \left | \begin{array}{cc}f & g \\ h^{'} & k \end{array} \right |$

(ii) $\left | \begin{array}{cc}f & g^{'} \\ h & k^{'} \end{array}\right | + \left | \begin{array}{cc} f^{'} & g \\ h & k^{'} \end{array} \right |$

(iii) $\left | \begin{array}{cc}f^{'} & g^{'} \\ h & k \end{array} \right | + \left | \begin{array}{cc}f & g \\ h^{'} & h^{'} \end{array} \right |$

(iv) $\left | \begin{array}{cc}f & g \\ h^{'} & k^{'} \end{array} \right | + \left | \begin{array}{cc}f^{'} & g \\h & k \end{array} \right |$

Problem 7:

If $pv=81$ , then $\frac{dp}{dv}$ at $v=9$ is equal to:

(i) 1 (ii) -1 (iii) 2 (iv) 3

Problem 8:

If $x^{2}+y^{2}=1$ , then

(i) $yy^{''}-2(y^{'})^{2}+1=0$ (ii) $yy^{''} - (y^{'})^{2}-1=0$ (iii) $yy^{''} + (y^{'})^{2} + 1 = 0$ (iv) $yy^{''} - 2(y^{'})^{2}-1=0$

Problem 9:

If $y = \arctan{\frac{\sqrt{x}-1}{\sqrt{x}+1}} + \arctan{\frac{\sqrt{x}+1}{\sqrt{x}-1}}$ , then the value of $\frac{dy}{dx}$ will be

(i) 0 (ii) 1 (iii) -1 (iv) $- \frac{1}{2}$

Problem 10:

Let $f(x) = \left | \begin{array}{ccc} x^{3} & \sin{x} & \cos{x} \\ 0 & -1 & 0 \\ p & p^{2} & p^{3} \end{array} \right |$ , where p is a constant, then $\frac{d^{3}}{dx^{3}}(f(x))$ at $x=0$ is

(a) p (b) $p+p^{2}$ (c) $p+p^{3}$ (d) independent of p

Regards,

Nalin Pithwa

By Nalin Pithwa | Also posted in calculus | Comments (0)

Derivatives: part 4: IITJEE maths tutorial problems for practice

November 13, 2020 – 11:16 pm

Problem 1:

Given $x=x(t)$ , $y=y(t)$ , then $\frac{d^{2}y}{dx^{2}}$ is equal to

(a) $\frac{\frac{d^{2}y}{dt^{2}}}{\frac{d^{2}x}{dt^{2}}}$

(b) $\frac{\frac{d^{2}y}{dt^{2}} \times \frac{dx}{dt} - \frac{dy}{dt} \times \frac{d^{2}x}{dt^{2}}}{(\frac{dx}{dt})^{3}}$

(c) $\frac{\frac{dx}{dt} \times \frac{d^{2}y}{dt^{2}} - \frac{d^{2}x}{dt^{2}} \times \frac{dy}{dt}}{(\frac{dx}{dt})^{2}}$

(d) $\frac{1}{\frac{d^{2}x}{dy^{2}}}$

Problem 2:

$\frac{d}{dx}(\arctan{\sec{x}+ \tan{x}})$ is equal to

(a) 0 (b) $\sec{x}-\tan{x}$ (c) $\frac{1}{2}$ (d) 2

Problem 3:

If $y= \sqrt{x + \sqrt{x + \sqrt{x} + \ldots}}$ , then $\frac{dy}{dx}$ is equal to :

(a) 1 (b) \ $\frac{1}{xy}$ (c) $\frac{1}{2y-x}$ (d) $\frac{1}{2y-1}$

Problem 4:

If $f(x) = \left| \begin{array}{ccc} x & x^{2} & x^{3} \\ 1 & 2x & 3x^{2} \\ 0 & 2 & 6x \end{array} \right|$ , then $f^{'}(x) =$

(a) 12 (b) $6x^{2}$ (c) $6x$ (d) $12x^{2}$

Problem 5:

If $y = (\frac{x^{a}}{x^{b}}) ^{a+b} \times (\frac{x^{b}}{x^{c}})^{b+c} \times (\frac{x^{c}}{x^{a}})^{c+a}$ , then $\frac{dy}{dx}=$

(a) 0 (b) 1 (c) $a+b+c$ (d) abc

Problem 6:

If $y = \arctan{\frac{x-\sqrt{1-x^{2}}}{x+\sqrt{1-x^{2}}}}$ , then $\frac{dy}{dx}$ is equal to

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

Problem 7:

If $x=at^{2}$ , $y=2at$ , then $\frac{d^{2}y}{dx^{2}}=$

(a) $\frac{1}{t^{2}}$ (b) $\frac{1}{2at^{3}}$ (c) $\frac{1}{t^{3}}$ (d) $\frac{-1}{2at^{3}}$

Problem 8:

If $y=ax^{n+1} +bx^{-n}$ , then $x^{2}\frac{d^{2}y}{dx^{2}}=$

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

Problem 9:

If $x=t^{2}$ , $y=t^{3}$ , then $\frac{d^{2}y}{dx^{2}}=$

(a) $\frac{3}{2}$ (b) $\frac{3}{4t}$ (c) $\frac{3}{2t}$ (d) 0

Problem 10:

If $y=a+bx^{2}$ , a, b arbitrary constants, then

(a) $\frac{d^{2}}{dx^{2}} = 2xy$ (b) $x \frac{d^{2}y}{dx^{2}} - \frac{dy}{dx} + y=0$ (c) $x \frac{d^{2}y}{dx^{2}} = \frac{dy}{dx}$ (d) $x \frac{d^{2}y}{dx^{2}} = 2xy$

Regards,

Nalin Pithwa

By Nalin Pithwa | Also posted in calculus | Comments (0)

Derivatives: part 3: IITJEE maths tutorial problems for practice

October 28, 2020 – 2:34 pm

Problem 1:

Differential coefficient of $\log[10]{x}$ w.r.t. $\log[x]{10}$ is

(a) $\frac{(\log{x})^{2}}{(\log{10})^{2}}$ (b) $\frac{(\log[x]{10})^{2}}{(\log{10})^{2}}$ (c) $\frac{(\log[10]{x})^{2}}{(\log{10})^{2}}$ (d) $\frac{(\log{10})^{2}}{(\log{x})^{2}}$

Problem 2:

The derivative of an even function is always:

(a) an odd function (b) does not exist (c) an even function (d) can be either even or odd.

Problem 3:

The derivative of $\arcsin{x}$ w.r.t. $\arccos{\sqrt{1-x^{2}}}$ is

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

Problem 4:

If $\sqrt{1-x^{2}}+\sqrt{1-y^{2}}=a(x-y)$ , then $\frac{dy}{dx}$ is

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

Problem 5:

$\frac{d}{dx} \arcsin{2x\sqrt{1-x^{2}}}$ is equal to

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

Problem 6:

If $y=\arctan{\frac{x}{2}}-\arccos{\frac{x}{2}}$ , then $\frac{dy}{dx}$ is

(a) $\frac{2}{1+x^{2}}$ (b) $\frac{2}{4+x^{2}}$ (c) $\frac{4}{4+x^{2}}$ (d) $0$

Problem 7:

If $y=\arccos{(\frac{\sqrt{1+\sin{x}}+\sqrt{1-\sin{x}}}{\sqrt{1+\sin{x}}-\sqrt{1-\sin{x}}})}$ , then $\frac{dy}{dx}$ is equal to:

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

Problem 8:

If $y = \arctan{\frac{4x}{1+5x^{2}}} + \arctan{\frac{2+3x}{3-2x}}$ , then $\frac{dy}{dx}$ is

(a) $\frac{1}{1+x^{2}}$ (b) $\frac{5}{1+25x^{2}}$ (c) $1$ (d) $\frac{3}{1+9x^{2}}$

Problem 9:

If $2^{x}+2^{y}=2^{x+y}$ , then $\frac{dy}{dx}$ is equal to

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