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}})

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.

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