Category Archives: calculus

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.

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<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

Derivatives: part 8: IITJEE mains tutorial problems practice

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)}}

(c) \frac{\sin{(a-b)}}{\sin{(x-a)}\sin{(x-b)}} (d) \frac{\cos{(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.

Derivatives: part 7: IITJEE tutorial problems practice

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})}

(c) \frac{1}{\sin{\sqrt{a-ax}}} (d) zero

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.

Derivatives: part 6: IITJEE tutorial practice problems

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

Derivatives: part 5: IITJEE maths tutorial problems for practice

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

Derivatives: part 4: IITJEE maths tutorial problems for practice

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

Derivatives: part 3: IITJEE maths tutorial problems for practice

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}}

Problem 10:

If y^{2}=p(x), a polynomial of degree 3, then 2\frac{d}{dx}(y^{3}\frac{d^{2}y}{dx^{2}}) is equal to

(a) p^{'''}(x)+p^{'}(x) (b) p^{''}(x).p^{'''}(x) (c) p^{'''}(x).p(x) (d) a constant.

Regards,

Nalin Pithwa.

Derivatives : part 2: IITJEE Maths : Tutorial problems for practice

Problem 1:

If f(a)=2, f^{'}(a)=1, g(a)=-1, g^{'}(a)=2, then the value of \lim_{x \rightarrow a}\frac{g(x)f(a)-g(a)f(x)}{x-a} is

(a) -5 (b) \frac{1}{5} (c) 5 (d) 0

Problem 2:

Let y = \arcsin{(\frac{2x}{1+x^{2}})}, 0 < x <1 and 0 < y < \frac{\pi}{2}, then \frac{dy}{dx} is equal to :

(a) \frac{2}{1+x^{2}} (b) \frac{2x}{1+x^{2}} (c) \frac{-2}{1+x^{2}} (d) none

Problem 3:

Let f(x) = ax^{2}+1 for x \leq 1

and f(x)= x+a for x \leq 1 then f is derivable at x=1, if

(a) a=0 (b) a = \frac{1}{2} (c) a=1 (d) a=2

Problem 4:

If f(x) = ax^{2}+b for x \leq 1

if f(x)=b x^{2}+ax+c for x>1, where b \neq 0, then f(x) is continuous and differentiable at x=1, if

(a) c=0, a=2b (b) a=2b, c \in \Re (c) a=b, c=0 (d) a=2b, c \neq 0

Problem 5:

\lim_{h \rightarrow 0} \frac{\cos^{2}(x+h)- \cos^{2}(x)}{h} is equal to

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

Problem 6:

\lim_{h \rightarrow 0} \frac{\sin{\sqrt{x+h}-\sin{\sqrt{x}}}}{h} is equal to

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

Problem 7:

(\arccos{x})^{'}= \frac{-1}{\sqrt{1-x^{2}}} where

(a) -1 < x <1 (b) -1 \leq x \leq 1 (c) -1 \leq x < 1 (d) -1 < x \leq 1

Problem 8:

\frac{d}{dx}(\arctan{(\frac{3x-x^{2}}{1-3x^{2}})}) is equal to

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

Problem 9:

If x=a\cos^{3}(t) and y=a\sin^{3}(t), then \frac{dy}{dx} is equal to

(a) \cos{t} (b) \cot{t} (c) cosec{(t)} (d) -\tan{t}

Problem 10:

If y = arcsin{\cos{x}}, then \frac{dy}{dx} is equal to

(a) -1 (b) \cos{t} (c) cosec{(t)} (d) -\tan{t}

Regards,

Nalin Pithwa