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:

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

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

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

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

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