Rd sharma solution class 12 chapter Differentiation

Exercise 11.2

Question 1

Differentiate 

sin~(3x+5)

Sol :

Let y=sin~(3x+5)

Differentiating y with respect to x we get,

\dfrac{dy}{dx}=\dfrac{d}{dx}\{sin~(3x+5)\}

=cos~(3x+5)\dfrac{d}{dx}(3x+5) [using~chain~rule]

=cos~(3x+5)\times 3

=3~cos~(3x+5)

So, \dfrac{d}{dx}\left\{sin~(3x+5)\right\}=3~cos~(3x+5)

 

 

Question 2

Differentiate

tan^2~x

Sol :

Let y=tan^2~x

which can be written as

y=(tan~x)^2

Differentiating y with respect to x we get,

\dfrac{dy}{dx}=\dfrac{d}{dx}(2~tan~x)\times\dfrac{d}{dx}(tan~x) [using~chain~rule]

=2~tan~x\times sec^2x

So, \dfrac{d}{dx}\left\{tan^2~x\right\}=2~tan~x~sec^2~x

 

 

Question 3

Differentiate

tan~(x^\circ+45^\circ)

Sol :

Let y=tan~(x^\circ+45^\circ)

\Rightarrow y=tan~\left\{(x+45)\dfrac{\pi}{180}\right\}

Differentiating y with respect to x we get,

\dfrac{dy}{dx}=\dfrac{d}{dx}tan~\left\{(x+45)\dfrac{\pi}{180}\right\} [using~chain~rule]

=sec^2~\left\{(x+45)\dfrac{\pi}{180}\right\}\times \dfrac{d}{dx}(x+45)\dfrac{\pi}{180}

=\dfrac{\pi}{180}sec^2~(x^\circ+45^\circ)

So, \dfrac{d}{dx}\left\{tan~(x^\circ+45^\circ)\right\}=\dfrac{\pi}{180}sec^2~(x^\circ+45^\circ)

 

Question 4

Differentiate

sin~(log~x)

Sol :

Let y=sin~(log~x)

Differentiating y with respect to x we get,

\dfrac{dy}{dx}=\dfrac{d}{dx}sin~(log~x)

=cos~(log~x)\times\dfrac{d}{dx}(log~x) [using~chain~rule]

=\dfrac{1}{x}cos~(log~x)

So, \dfrac{d}{dx}\left\{sin~(log~x)\right\}=\dfrac{1}{x}cos~(log~x)

 

 

Question 5

Differentiate

e^{sin~\sqrt x}

Sol :

Let y=e^{sin~\sqrt x}

Differentiating y with respect to x we get,

\dfrac{dy}{dx}=\dfrac{d}{dx}\big(e^{sin~\sqrt x}\big)

=e^{sin~\sqrt x}\times\dfrac{d}{dx} (sin~\sqrt{x}) [using~chain~rule]

=e^{sin~\sqrt x}\times cos~\sqrt{x}\times\dfrac{d}{dx}(\sqrt{x}) [using~chain~rule]

=e^{sin~\sqrt x}\times cos~\sqrt{x}\times \dfrac{1}{2\sqrt{x}}

=\dfrac{cos~\sqrt{x}~e^{sin~\sqrt x}}{2\sqrt{x}}

So, \dfrac{d}{dx}(e^{sin~\sqrt{x}})=\dfrac{cos~\sqrt{x}~e^{sin~\sqrt x}}{2\sqrt{x}}

 

 

Question 6

Differentiate

e^{tan~x}

Sol :

Let y=e^{tan~x}

Differentiating y with respect to x we get,

\dfrac{dy}{dx}=\dfrac{d}{dx}(e^{tan~x})

=e^{tan~x}\times\dfrac{d}{dx}~(tan~x) [using~chain~rule]

=e^{tan~x}\times sec^2~x

=sec^2~x\times e^{tan~x}

So, \dfrac{d}{dx}(e^{tan~x})=sec^2~x\times e^{tan~x}

 

 

Question 7

Differentiate

sin^2~(2x+1)

Sol :

Let y=sin^2~(2x+1)

which can be written as

y=\left\{sin~(2x+1)^2\right\}

Differentiating y with respect to x we get,

\dfrac{dy}{dx}=\dfrac{d}{dx}\bigg\{2~sin~(2x+1)\bigg\}\times\dfrac{d}{dx}\bigg\{sin~(2x+1)\bigg\} [using~chain~rule]

=2~sin~(2x+1)~.~cos~(2x+1)\times\dfrac{d}{dx}(2x+1) [using~chain~rule]

=2~sin~(2x+1)~cos~(2x+1)\times 2

=2~sin~2(2x+1) [\because 2~sin~x~cos~x=sin2x]

=2~sin~(4x+2)

So, \dfrac{d}{dx}\left\{sin^2~(2x+1)\right\}=2~sin~(4x+2)

 

 

Question 8

Differentiate

log_7~(2x+3)

Sol :

Let y=log_7~(2x+3)

Let \Rightarrow y=\dfrac{log~(2x+3)}{log_e~7} \bigg[\because log_a~b=\dfrac{log~b}{log~a}\bigg]

Differentiating y with respect to x we get,

\dfrac{dy}{dx}=\dfrac{d}{dx}\left\{log~(2x+3)\times \dfrac{1}{log_e~7}\right\}

=log~(2x+3)\times \dfrac{d}{dx}\bigg(\dfrac{1}{log_e~7}\bigg)+\dfrac{1}{log_e~7}\times \dfrac{d}{dx}\bigg(log~(2x+3)\bigg) [using~product~rule]

=0+\dfrac{1}{log_e~7}\times \dfrac{d}{dx}\bigg(log~(2x+3)\bigg)

=\dfrac{1}{log_e~7}\times \dfrac{1}{2x+3}\times \dfrac{d}{dx}(2x+3)

=\dfrac{2}{log_e~7~(2x+3)}

So, \dfrac{d}{dx}\left\{log_7~(2x+3)\right\}=\dfrac{2}{log_e~7~(2x+3)}

 

 

Question 9

Differentiate

tan~5x^\circ

Sol :

Let y=tan~5x^\circ

\Rightarrow y=tan~5x~\dfrac{\pi}{180}

Differentiating y with respect to x we get,

\dfrac{dy}{dx}=\dfrac{d}{dx}\bigg(tan~5x~\dfrac{\pi}{180}\bigg)

=sec^2\bigg(5x\times\dfrac{\pi}{180}\bigg)\times\dfrac{d}{dx}\bigg(5x\times\dfrac{\pi}{180}\bigg) [using~chain~rule]

=sec^2\bigg(5x\times\dfrac{\pi}{180}\bigg)\times\bigg\{5x\times\dfrac{d}{dx}\bigg(\dfrac{\pi}{180}\bigg)+\dfrac{\pi}{180}\times\dfrac{d}{dx}(5x)\bigg\} [using~chain~rule]

=sec^2\bigg(5x\times\dfrac{\pi}{180}\bigg)\times\bigg\{0+\dfrac{\pi}{180}\times\dfrac{d}{dx}(5x)\bigg\} [using~product~rule]

=sec^2\bigg(5x\times\dfrac{\pi}{180}\bigg)\times\bigg\{\dfrac{\pi}{180}\times5\bigg\}

=sec^2\bigg(5x\times\dfrac{\pi}{180}\bigg)\times\bigg\{\dfrac{5\pi}{180}\bigg\}

=\dfrac{5\pi}{180}sec^2~(5x^\circ)

So, \dfrac{d}{dx}\left\{tan~5x^\circ\right\}=\dfrac{5\pi}{180}sec^2~(5x^\circ)

 

 

Question 10

Differentiate

2^{x^3}

Sol :

Let y=2^{x^3}

Differentiating y with respect to x we get,

\dfrac{dy}{dx}=\dfrac{d}{dx}2^{x^3}

=2^{x^3}~log_e\times2\times\dfrac{d}{dx}(x^3) [using~chain~rule]

=2^{x^3}~log_e\times2\times3x^2

So, \dfrac{d}{dx}\left\{2^{x^3}\right\}=2^{x^3}~log_e\times2\times3x^2

 

 

Question 11

Differentiate

3^{e^x}

Sol :

Let y=3^{e^x}

Differentiating y with respect to x we get,

\dfrac{dy}{dx}=\dfrac{d}{dx}(3^{e^x})

=3^{e^x}\times\dfrac{d}{dx}(e^x) [using~chain~rule]

=e^x\times3^{e^x}

So, \dfrac{d}{dx}\left\{3^{e^x}\right\}=e^x\times3^{e^x}

 

 

Question 12

Differentiate

log_x~3

Sol :

Let y=log_x~3

\Rightarrow y=\dfrac{log~3}{log~x} \bigg[\because log_ab=\dfrac{log~b}{log~a}\bigg]

Differentiating y with respect to x we get,

\dfrac{dy}{dx}=\bigg\{log~3\times\dfrac{d}{dx}\bigg(\dfrac{1}{log~x}\bigg)+\dfrac{1}{log~x}\times\dfrac{d}{dx}(log~3)\bigg\} [using~product~rule]

=\bigg\{log~3\times\dfrac{d}{dx}\bigg(\dfrac{1}{log~x}\bigg)+0\bigg\}

=\bigg\{log~3\times\dfrac{d}{dx}(log~x)^{-1}\bigg\}

=log~3\times-1(log~x)^{-2}\times\dfrac{d}{dx}(log~x)

=log~3\times-1(log~x)^{-2}\times\dfrac{1}{x}

=-\dfrac{log~3}{(log~x)^{2}}\times\dfrac{1}{x}

=-\dfrac{(log~3)^2}{(log~x)^{2}}\times\dfrac{1}{x}\times\dfrac{1}{log~3}

=-\dfrac{1}{x~log~3(log~x)^{2}} \because \dfrac{log~b}{log~a}=\dfrac{1}{log_b~a}

So, \dfrac{d}{dx}\left\{log_x~3\right\}=-\dfrac{1}{x~log~3(log~x)^{2}}

 

Question 13

Differentiate

3^{x^2+2x}

Sol :

Let y=3^{x^2+2x}

Differentiating y with respect to x we get,

\dfrac{dy}{dx}=\dfrac{d}{dx}3^{x^2+2x}

=3^{x^2+2x}\times log_e3\times\dfrac{d}{dx}(x^2+2x) [using~chain~rule]

=3^{x^2+2x}\times log_e3\times\bigg\{2x+\dfrac{d}{dx}(2x)\bigg\}

=3^{x^2+2x}\times log_e3\times(2x+2)

=(2x+2)3^{x^2+2x}{.}~log_e3

So, \dfrac{d}{dx}\left\{3^{x^2+2x}\right\}=(2x+2)3^{x^2+2x}{.}~log_e3

 

 

Question 14

Differentiate

\sqrt{\dfrac{a^2-x^2}{a^2+x^2}}

Sol :

Let y=\sqrt{\dfrac{a^2-x^2}{a^2+x^2}}

Differentiating y with respect to x we get,

\dfrac{dy}{dx}=\dfrac{d}{dx}\bigg(\dfrac{a^2-x^2}{a^2+x^2}\bigg)^\frac{1}{2}

=\dfrac{1}{2}\bigg(\dfrac{a^2-x^2}{a^2+x^2}\bigg)^{\frac{1}{2}-1}\times\dfrac{d}{dx}\bigg(\dfrac{a^2-x^2}{a^2+x^2}\bigg) [using~chain~rule]

=\dfrac{1}{2}\bigg(\dfrac{a^2-x^2}{a^2+x^2}\bigg)^\frac{-1}{~2}\times\left\{\dfrac{(a^2+x^2)\dfrac{d}{dx}(a^2-x^2)-(a^2-x^2)\dfrac{d}{dx}(a^2+x^2)}{(a^2+x^2)^2}\right\} [using~quotien~rule]

=\dfrac{1}{2}\bigg(\dfrac{a^2-x^2}{a^2+x^2}\bigg)^\frac{-1}{~2}\times\left\{\dfrac{(a^2+x^2)(-2x)-(a^2-x^2)(2x)}{(a^2+x^2)^2}\right\}

=\dfrac{1}{2}\bigg(\dfrac{a^2-x^2}{a^2+x^2}\bigg)^\frac{-1}{~2}\times\left\{\dfrac{-2xa^2-2x^3-2xa^2+2x^3}{(a^2+x^2)^2}\right\}

=\dfrac{1}{2}\bigg(\dfrac{a^2-x^2}{a^2+x^2}\bigg)^\frac{-1}{~2}\times\left\{\dfrac{-4xa^2}{(a^2+x^2)^2}\right\}

=\dfrac{1}{2}\bigg(\dfrac{a^2+x^2}{a^2-x^2}\bigg)^\frac{1}{~2}\times\dfrac{-4xa^2}{(a^2+x^2)^2}

=\dfrac{(a^2+x^2)^\frac{1}{~2}}{(a^2-x^2)^\frac{1}{~2}}\times\dfrac{-2xa^2}{(a^2+x^2)^2}

=\dfrac{-2xa^2}{(a^2-x^2)^\frac{1}{~2}(a^2+x^2)^2(a^2+x^2)^\frac{-1}{~2}}

=\dfrac{-2xa^2}{\sqrt{a^2-x^2}(a^2+x^2)^\frac{3}{~2}}

So, \dfrac{d}{dx}\left\{\sqrt{\dfrac{a^2-x^2}{a^2+x^2}}\right\}=\dfrac{-2xa^2}{\sqrt{a^2-x^2}(a^2+x^2)^\frac{3}{~2}}

 

Question 15

Differentiate

3^{x~log~x}

Sol :

Let y=3^{x~log~x}

Differentiating y with respect to x we get,

\dfrac{dy}{dx}=\dfrac{d}{dx}3^{x~log~x}

=3^{x~log~x}\times log_e3\times\dfrac{d}{dx}(x~log~x) [using~chain~rule]

=3^{x~log~x}\times log_e3\times\left\{(x)\dfrac{d}{dx}(log~x)+(log~x)\dfrac{d}{dx}(x)\right\} [using~product~rule]

=3^{x~log~x}\times log_e3\times\left\{\dfrac{x}{x}+log~x\right\}

=3^{x~log~x}\times log_e3\times (1+log~x)

So, \dfrac{d}{dx}\left\{3^{x~log~x}\right\}=3^{x~log~x}\times log_e3\times (1+log~x)

 

Question 16

Differentiate

\sqrt{\dfrac{1+sin~x}{1-sin~x}}

Sol :

 

Let y=\sqrt{\dfrac{1+sin~x}{1-sin~x}}

Differentiating y with respect to x we get,

\dfrac{dy}{dx}=\dfrac{d}{dx}\bigg(\dfrac{1+sin~x}{1-sin~x}\bigg)^\frac{1}{2}

=\dfrac{1}{2}\bigg(\dfrac{1+sin~x}{1-sin~x}\bigg)^{\frac{1}{2}-1}\times\dfrac{d}{dx}\bigg(\dfrac{1+sin~x}{1-sin~x}\bigg) [using~chain~rule]

=\dfrac{1}{2}\bigg(\dfrac{1+sin~x}{1-sin~x}\bigg)^{\frac{1}{2}-1}\times\left\{\dfrac{(1-sin~x)\dfrac{d}{dx}(1+sin~x)-(1+sin~x)\dfrac{d}{dx}(1-sin~x)}{(1-sin~x)^2}\right\} [using~quotient~rule]

=\dfrac{1}{2}\bigg(\dfrac{1+sin~x}{1-sin~x}\bigg)^{\frac{1}{2}-1}\times\left\{\dfrac{(1-sin~x)(cos~x)-(1+sin~x)(-cos~x)}{(1-sin~x)^2}\right\}

=\dfrac{1}{2}\bigg(\dfrac{1+sin~x}{1-sin~x}\bigg)^{\frac{-1}{~2}}\times\left\{\dfrac{cos~x-sin~x~cos~x+cos~x+sin~x~cos~x}{(1-sin~x)^2}\right\}

=\dfrac{1}{2}\bigg(\dfrac{1-sin~x}{1+sin~x}\bigg)^{\frac{1}{~2}}\times\left\{\dfrac{2~cos~x}{(1-sin~x)^2}\right\}

=\dfrac{(1-sin~x)^{\frac{1}{~2}}}{(1+sin~x)^{\frac{1}{~2}}}\times\dfrac{cos~x}{(1-sin~x)^2}

=\dfrac{cos~x}{\sqrt{1+sin~x}~(1-sin~x)^2(1-sin~x)^{\frac{-1}{~2}}}

=\dfrac{cos~x}{\sqrt{1+sin~x}~(1-sin~x)^{\frac{3}{2}}}

=\dfrac{cos~x}{\sqrt{1+sin~x}~\sqrt{1-sin~x}(1-sin~x)}

=\dfrac{cos~x}{\sqrt{1-sin^2~x}~(1-sin~x)}

=\dfrac{cos~x}{\sqrt{cos^2~x}~(1-sin~x)} [\because 1-sin^2~x=cos~^2~x]

=\dfrac{1}{(1-sin~x)}\times\dfrac{(1+sin~x)}{(1+sin~x)} on rationalising

=\dfrac{1+sin~x}{cos^2~x}

=\dfrac{1}{cos~x}\bigg(\dfrac{1}{cos~x}+\dfrac{sin~x}{cos~x}\bigg)

=sec~x\big(sec~x+tan~x\big)

=sec^2~x+tan~x~sec~x

So, \dfrac{d}{dx}\left\{\sqrt{\dfrac{1+sin~x}{1-sin~x}}\right\}=sec^2~x+tan~x~sec~x

 

 

Question 17

Differentiate

\sqrt{\dfrac{1-x^2}{1+x^2}}

Sol :

Let y=\sqrt{\dfrac{1-x^2}{1+x^2}}

Differentiating y with respect to x we get,

\dfrac{dy}{dx}=\dfrac{d}{dx}\bigg(\dfrac{1-x^2}{1+x^2}\bigg)^\frac{1}{2}

=\dfrac{1}{2}\bigg(\dfrac{1-x^2}{1+x^2}\bigg)^{\frac{1}{2}-1}\dfrac{d}{dx}\bigg(\dfrac{1-x^2}{1+x^2}\bigg) [using~chain~rule]

=\dfrac{1}{2}\bigg(\dfrac{1-x^2}{1+x^2}\bigg)^{\frac{-1}{~2}}\times\left\{\bigg(\dfrac{(1+x^2)\dfrac{d}{dx}(1-x^2)-(1-x^2)\dfrac{d}{dx}(1+x^2)}{(1+x^2)^2}\right\} [using~quotient~rule]

=\dfrac{1}{2}\bigg(\dfrac{1-x^2}{1+x^2}\bigg)^{\frac{-1}{~2}}\times\left\{\dfrac{(1+x^2)(2x)-(1-x^2)(2x)}{(1+x^2)^2}\right\}

=\dfrac{1}{2}\bigg(\dfrac{1+x^2}{1-x^2}\bigg)^{\frac{1}{2}}\times\left\{\dfrac{(2x+2x^3)-(2x-2x^3)}{(1+x^2)^2}\right\}

=\dfrac{1}{2}\dfrac{(1+x^2)^{\frac{1}{2}}}{(1-x^2)^{\frac{1}{2}}}\times\dfrac{4x^3}{(1+x^2)^2}

=\dfrac{2x^3}{\sqrt{1-x^2}~(1+x^2)^2(1+x^2)^{\frac{-1}{~2}}}

=\dfrac{2x^3}{\sqrt{1-x^2}~(1+x^2)^{\frac{3}{2}}}

So, \dfrac{d}{dx}\left\{\sqrt{\dfrac{1-x^2}{1+x^2}}\right\}=\dfrac{2x^3}{\sqrt{1-x^2}~(1+x^2)^{\frac{3}{2}}}

 

 

Question 18

Differentiate

(log~sin~x)^2

Sol :

Let y=(log~sin~x)^2

Differentiating y with respect to x we get,

\dfrac{dy}{dx}=\dfrac{d}{dx}(log~sin~x)^2

=2~(log~sin~x)\times\dfrac{d}{dx}(log~sin~x) [using~chain~rule]

=2~(log~sin~x)\times\dfrac{1}{sin~x}\times\dfrac{d}{dx}(sin~x)

=2~(log~sin~x)\times\dfrac{1}{sin~x}\times(cos~x)

=2~(log~sin~x)\times cot~x

So, \dfrac{d}{dx}\left\{(log~sin~x)^2\right\}=2~(log~sin~x)~{.}~cot~x

 

 

Question 19

Differentiate

\sqrt{\dfrac{1+x}{1-x}}

Sol :

Let y=\sqrt{\dfrac{1+x}{1-x}}

Differentiating y with respect to x we get,

\dfrac{dy}{dx}=\dfrac{d}{dx}\bigg(\dfrac{1+x}{1-x}\bigg)^\frac{1}{2}

=\dfrac{1}{2}\times\bigg(\dfrac{1+x}{1-x}\bigg)^{\frac{1}{2}-1}\times\dfrac{d}{dx}\bigg(\dfrac{1+x}{1-x}\bigg) [using~chain~rule]

=\dfrac{1}{2}\times\bigg(\dfrac{1+x}{1-x}\bigg)^{\frac{1}{2}-1}\times\left\{\dfrac{(1-x)\dfrac{d}{dx}(1+x)-(1+x)\dfrac{d}{dx}(1-x)}{(1-x)^2}\right\} [using~product~rule]

=\dfrac{1}{2}\times\bigg(\dfrac{1+x}{1-x}\bigg)^{\frac{-1}{~2}}\times\left\{\dfrac{1-x+1+x}{(1-x)^2}\right\} 

=\dfrac{1}{2}\times\bigg(\dfrac{1-x}{1+x}\bigg)^{\frac{1}{2}}\times\dfrac{2}{(1-x)^2}

=\dfrac{(1-x)^{\frac{1}{2}}}{(1+x)^{\frac{1}{2}}}\times\dfrac{1}{(1-x)^2}

=\dfrac{-x}{\sqrt{1+x}(1-x)^{\frac{-1}{~2}}(1-x)^2}

=\dfrac{1}{\sqrt{1+x}~(1-x)^{\frac{3}{2}}}

So, \dfrac{d}{dx}\left\{\sqrt{\dfrac{1+x}{1-x}}\right\}=\dfrac{1}{\sqrt{1+x}~(1-x)^{\frac{3}{2}}}

 

Question 20

Differentiate

sin~\bigg(\dfrac{1+x^2}{1-x^2}\bigg)

Sol :

Let y=sin~\bigg(\dfrac{1+x^2}{1-x^2}\bigg)

Differentiating y with respect to x we get,

\dfrac{dy}{dx}=\dfrac{d}{dx}\bigg[sin~\bigg(\dfrac{1+x^2}{1-x^2}\bigg)\bigg]

=cos~\bigg(\dfrac{1+x^2}{1-x^2}\bigg)\times\dfrac{d}{dx}\bigg(\dfrac{1+x^2}{1-x^2}\bigg) [using~chain~rule]

=cos~\bigg(\dfrac{1+x^2}{1-x^2}\bigg)\times\left\{\dfrac{(1-x^2)\dfrac{d}{dx}(1+x^2)-(1+x^2)\dfrac{d}{dx}(1-x^2)}{(1-x^2)^2}\right\} [using~quotient~rule]

=cos~\bigg(\dfrac{1+x^2}{1-x^2}\bigg)\times\left\{\dfrac{(1-x^2)(2x)-(1+x^2)(-2x)}{(1-x^2)^2}\right\}

=cos~\bigg(\dfrac{1+x^2}{1-x^2}\bigg)\times\left\{\dfrac{(2x-2x^3)+(2x+2x^3)}{(1-x^2)^2}\right\}

=cos~\bigg(\dfrac{1+x^2}{1-x^2}\bigg)\times\dfrac{4x}{(1-x^2)^2}

=\dfrac{4x}{(1-x^2)^2}\times cos~\bigg(\dfrac{1+x^2}{1-x^2}\bigg)

So, \dfrac{d}{dx}\left\{sin~\bigg(\dfrac{1+x^2}{1-x^2}\bigg)\right\}=\dfrac{4x}{(1-x^2)^2}\times cos~\bigg(\dfrac{1+x^2}{1-x^2}\bigg)

 

Question 21

Differentiate

e^{3x}~cos~2x

Sol :

Let y=e^{3x}~cos~2x

Differentiating y with respect to x we get,

\dfrac{dy}{dx}=\dfrac{d}{dx}(e^{3x}~cos~2x)

=\left\{e^{3x}\times\dfrac{d}{dx}(cos~2x)+(cos~2x)\times\dfrac{d}{dx}(e^{3x})\right\} [using~product~rule]

=\left\{e^{3x}\times(-sin~2x)(2)+(cos~2x)\times(e^{3x})(3)\right\} [using~chain~rule]

=-2~sin~2x~e^{3x}+3~e^{3x}~cos~2x

=e^{3x}(-2~sin~2x+3~cos~2x)

So, \dfrac{d}{dx}\left\{e^{3x}~cos~2x\right\}=e^{3x}(-2~sin~2x+3~cos~2x)

 

 

 

Question 22

Differentiate

sin~(log~sin~x)

Sol :

Let y=sin~(log~sin~x)

Differentiating y with respect to x we get,

\dfrac{dy}{dx}=\dfrac{d}{dx}sin~(log~sin~x)

=cos~(log~sin~x)\times\dfrac{d}{dx}(log~sin~x) [using~chain~rule]

=cos~(log~sin~x)\times\dfrac{1}{sin~x}\times(cos~x)

=cos~(log~sin~x)\times(cot~x)

So, \dfrac{d}{dx}\left\{sin~(log~sin~x)\right\}=cos~(log~sin~x)\times(cot~x)

 

 

Question 23

Differentiate

e^{tan~3x}

Sol :

Let y=e^{tan~3x}

Differentiating y with respect to x we get,

\dfrac{dy}{dx}=\dfrac{d}{dx}(e^{tan~3x})

=e^{tan~3x}\times \dfrac{d}{dx}(tan~3x) [using~chain~rule]

=e^{tan~3x}\times (sec^2~x)\times \dfrac{d}{dx}(3x)

=3\times e^{tan~3x}\times (sec^2~x)

So, \dfrac{d}{dx}\left\{e^{tan~3x}\right\}=3~e^{tan~3x}~{.}~(sec^2~x)

 

 

Question 24

Differentiate

e^{\sqrt{cot~x}}

Sol :

Let y=e^{\sqrt{cot~x}}

Differentiating y with respect to x we get,

\dfrac{dy}{dx}=\dfrac{d}{dx}(e^{\sqrt{cot~x}})

=e^{\sqrt{cot~x}}\times\dfrac{d}{dx}(cot~x)^{\frac{1}{2}} [using~chain~rule]

=e^{\sqrt{cot~x}}\times\dfrac{1}{2}(cot~x)^{\frac{1}{2}-1}\times\dfrac{d}{dx}(cot~x) [using~chain~rule]

=e^{\sqrt{cot~x}}\times\dfrac{1}{2}(cot~x)^{\frac{-1}{~2}}\times-(cosec^2~x)

=-\dfrac{(cosec^2~x)\times e^{\sqrt{cot~x}}}{2\sqrt{cot~x}}

So, \dfrac{d}{dx}\left\{e^{\sqrt{cot~x}}\right\}=-\dfrac{(cosec^2~x)\times e^{\sqrt{cot~x}}}{2\sqrt{cot~x}}

 

Question 25

Differentiate

log~\dfrac{sin~x}{1+cos~x}

Sol :

Let y=log~\dfrac{sin~x}{1+cos~x}

Differentiating y with respect to x we get,

\dfrac{dy}{dx}=\dfrac{d}{dx}\bigg\{log~\bigg(\dfrac{sin~x}{1+cos~x}\bigg)\bigg\}

=\dfrac{1}{\bigg(\dfrac{sin~x}{1+cos~x}\bigg)}\times\dfrac{d}{dx}\bigg(\dfrac{sin~x}{1+cos~x}\bigg) using~chain~rule

=\dfrac{1}{\bigg(\dfrac{sin~x}{1+cos~x}\bigg)}\times\left\{\dfrac{(1+cos~x)\dfrac{d}{dx}(sin~x)-(sin~x)\dfrac{d}{dx}(1+cos~x)}{(1+cos~x)^2}\right\} using~product~rule

=\dfrac{1}{\bigg(\dfrac{sin~x}{1+cos~x}\bigg)}\times\left\{\dfrac{cos~x+cos^2~x+sin^2~x}{(1+cos~x)^2}\right\}

=\dfrac{1}{\bigg(\dfrac{sin~x}{1+cos~x}\bigg)}\times\left\{\dfrac{cos~x+1}{(1+cos~x)^2}\right\} \because cos^2~x+sin^2~x=1

=\dfrac{1+cos~x}{sin~x}\times\dfrac{1}{1+cos~x}

=\dfrac{1}{sin~x}

=cosec~x

So, \dfrac{d}{dx}\left\{log~\dfrac{sin~x}{1+cos~x}\right\}=cosec~x

 

 

Question 26

Differentiate

log~\dfrac{1-cos~x}{1+cos~x}

Sol :

Let y=log~\bigg(\dfrac{1-cos~x}{1+cos~x}\bigg)

Differentiating y with respect to x we get,

\dfrac{dy}{dx}=\dfrac{d}{dx}\bigg\{log~\bigg(\dfrac{1-cos~x}{1+cos~x}\bigg)\bigg\}

=log~\bigg(\dfrac{1-cos~x}{1+cos~x}\bigg)\times\dfrac{d}{dx}\bigg(\dfrac{sin~x}{1+cos~x}\bigg) using~chain~rule

=\dfrac{1+cos~x}{1-cos~x}\times\dfrac{d}{dx}\bigg(\dfrac{1-cos~x}{1+cos~x}\bigg) using~chain~rule

=\dfrac{1+cos~x}{1-cos~x}\times\left\{\dfrac{(1+cos~x)\dfrac{d}{dx}(1-cos~x)-(1-cos~x)\dfrac{d}{dx}(1+cos~x)}{(1+cos~x)^2}\right\} using~product~rule

=\dfrac{1+cos~x}{1-cos~x}\times\left\{\dfrac{(1+cos~x)(sin~x)-(1-cos~x)(-sin~x)}{(1+cos~x)^2}\right\} 

=\dfrac{1+cos~x}{1-cos~x}\times\left\{\dfrac{sin~x+sin~x~cos~x+sin~x-sin~x~cos~x}{(1+cos~x)^2}\right\} 

=\dfrac{1}{(1-cos~x)}\times\dfrac{2~sin~x}{(1+cos~x)} 

=\dfrac{2~sin~x}{(1-cos^2~x)} 1-cos^2~x=sin^2~x

=\dfrac{2~sin~x}{sin^2~x} 

=2~cosec~x 

So, \dfrac{d}{dx}\left\{log~\dfrac{1-cos~x}{1+cos~x}\right\}=2~cosec~x

 

 

Question 27

Differentiate

tan~(e^{sin~x})

Sol :

Let y=tan~(e^{sin~x})

Differentiating y with respect to x we get,

\dfrac{dy}{dx}=\dfrac{d}{dx}\bigg\{tan~(e^{sin~x})\bigg\}

=sec^2~(e^{sin~x})\times \dfrac{d}{dx}(e^{sin~x}) [using~chain~rule]

=sec^2~(e^{sin~x})\times (e^{sin~x})\times\dfrac{d}{dx}({sin~x})

={sec}^2~(e^{sin~x})\times (e^{sin~x})\times({cos~x})

=(cos~x)~{.}~{sec}^2~(e^{sin~x})~{.}~(e^{sin~x})

So, \dfrac{d}{dx}\left\{tan~(e^{sin~x})\right\}=(cos~x)~{.}~{sec}^2~(e^{sin~x})~{.}~(e^{sin~x})

 

 

 

Question 28

Differentiate

log~(x+\sqrt{x^2+1})

Sol :

Let y=log~(x+\sqrt{x^2+1})

Differentiating y with respect to x we get,

\dfrac{dy}{dx}=\dfrac{d}{dx}\bigg\{log~(x+\sqrt{x^2+1})\bigg\}

=\dfrac{1}{x+\sqrt{x^2+1}}\times\dfrac{d}{dx}(x+\sqrt{x^2+1}) using~chain~rule

=\dfrac{1}{x+\sqrt{x^2+1}}\times1+\dfrac{1}{2\sqrt{x^2+1}}\times\dfrac{d}{dx}(x^2+1)

=\dfrac{1}{x+\sqrt{x^2+1}}\times 1+\dfrac{1}{2\sqrt{x^2+1}}\times 2x

=\dfrac{1}{x+\sqrt{x^2+1}}\times \dfrac{\sqrt{x^2+1}+~x}{\sqrt{x^2+1}}

=\dfrac{1}{\sqrt{x^2+1}}

So, \dfrac{d}{dx}\left\{log~(x+\sqrt{x^2+1})\right\}=\dfrac{1}{\sqrt{x^2+1}}

 

Question 29

Differentiate

\dfrac{e^x~log~x}{x^2}

Sol :

Let y=\dfrac{e^x~log~x}{x^2}

Differentiating y with respect to x we get,

\dfrac{dy}{dx}=\dfrac{d}{dx}\bigg\{\dfrac{e^x~log~x}{x^2}\bigg\}

=\left\{\dfrac{(x^2)\dfrac{d}{dx}(e^x~log~x)-(e^x~log~x)\dfrac{d}{dx}(x^2)}{(x^2)^2}\right\} using~quotient~rule

=\left[\dfrac{(x^2)\left\{(e^x)\dfrac{d}{dx}(~log~x)+(~log~x)\dfrac{d}{dx}(e^x)\right\}-(e^x~log~x)(2x)}{(x^2)^2}\right]

=\left[\dfrac{(x^2)\left\{\dfrac{e^x}{x}+(~log~x)(e^x)\right\}-(e^x~log~x)(2x)}{(x^2)^2}\right]

=e^x\left[\dfrac{(x^2)\left\{\dfrac{1}{x}+log~x\right\}-(2x~log~x)}{(x^2)^2}\right] taking~e^x~common

=e^x\left[\dfrac{(x^2)\left\{\dfrac{1+x~log~x}{x}\right\}-(2x~log~x)}{(x^2)^2}\right]

=e^x\left[\dfrac{(x)\left(1+x~log~x\right)-2x~log~x}{(x^2)^2}\right]

=x~e^x\left[\dfrac{1+x~log~x-2~log~x}{x^4}\right]

=e^x\left[\dfrac{1+x~log~x-2~log~x}{x^3}\right]

=e^x\left[\dfrac{1+x~log~x-2~log~x}{x^2.x}\right]

=x^{-2}.e^x\left[\dfrac{1+x~log~x-2~log~x}{x}\right]

=x^{-2}.e^x\bigg(\dfrac{1}{x}+\dfrac{x~log~x}{x}-\dfrac{2~log~x}{x}\bigg)

=x^{-2}.e^x\bigg(\dfrac{1}{x}+\dfrac{log~x}{x}-\dfrac{2~log~x}{x}\bigg)

 

So, \dfrac{d}{dx}\left\{\dfrac{e^x~log~x}{x^2}\right\}=x^{-2}.e^x\bigg(\dfrac{1}{x}+\dfrac{log~x}{x}-\dfrac{2~log~x}{x}\bigg)

 

Question 30

Differentiate

log~(cosec~x-cot~x)

Sol :

Let y=log~(cosec~x-cot~x)

Differentiating y with respect to x we get,

\dfrac{dy}{dx}=\dfrac{d}{dx}(log~(cosec~x-cot~x))

=\dfrac{1}{cosec~x-cot~x}\times\dfrac{d}{dx}(cosec~x-cot~x) using~chain~rule

=\dfrac{-cosec~x~cot~x+cosec^2~x}{cosec~x-cot~x}

=cosec~x\bigg(\dfrac{cosec~x-cot~x}{cosec~x-cot~x}\bigg) taking cosec x common

= cosec x

So, \dfrac{d}{dx}\left\{log~(cosec~x-cot~x)\right\}=cosec~x

 

 

Question 31

Differentiate

\dfrac{e^{2x}+e^{-2x}}{e^{2x}-e^{-2x}}

Sol :

Let y=\dfrac{e^{2x}+e^{-2x}}{e^{2x}-e^{-2x}}

Differentiating y with respect to x we get,

\dfrac{dy}{dx}=\dfrac{d}{dx}(\dfrac{e^{2x}+e^{-2x}}{e^{2x}-e^{-2x}})

=\left\{\dfrac{(e^{2x}-e^{-2x})\dfrac{d}{dx}(e^{2x}+e^{-2x})-(e^{2x}+e^{-2x})\dfrac{d}{dx}(e^{2x}-e^{-2x})}{(e^{2x}-e^{-2x})^2}\right\} using~quotient~rule

=\left\{\dfrac{(e^{2x}-e^{-2x})(2e^{2x}-2e^{-2x})-(e^{2x}+e^{-2x})(2e^{2x}+2e^{-2x})}{(e^{2x}-e^{-2x})^2}\right\}

=\left\{\dfrac{\{(2e^{2x})e^{2x}-(2e^{2x})e^{-2x}+e^{2x}(-2e^{-2x})-e^{-2x}(-2e^{-2x})\}-\{(2e^{2x})e^{2x}+(2e^{2x})e^{-2x}+(2e^{-2x})e^{2x}+(2e^{-2x})e^{-2x})\}}{(e^{2x}-e^{-2x})^2}\right\}

=\left\{\dfrac{\{2e^{4x}-2-2+2e^{-4x}\}-\{2e^{4x}+2+2+2e^{-4x}\}}{(e^{2x}-e^{-2x})^2}\right\}

=\left\{\dfrac{2e^{4x}-4+2e^{-4x}-2e^{4x}-4-2e^{-4x}}{(e^{2x}-e^{-2x})^2}\right\}

=\dfrac{-8}{(e^{2x}-e^{-2x})^2}

So, \dfrac{d}{dx}\left\{\dfrac{e^{2x}+e^{-2x}}{e^{2x}-e^{-2x}}\right\}=\dfrac{-8}{(e^{2x}-e^{-2x})^2}

 

 

Question 32

Differentiate

log~\bigg(\dfrac{x^2+x+1}{x^2-x+1}\bigg)

Sol :

Let y=log~\bigg(\dfrac{x^2+x+1}{x^2-x+1}\bigg)

Differentiating y with respect to x we get,

\dfrac{dy}{dx}=\dfrac{d}{dx}\left\{log~\bigg(\dfrac{x^2+x+1}{x^2-x+1}\bigg)\right\}

=\dfrac{x^2-x+1}{x^2+x+1}\times\dfrac{d}{dx}\bigg(\dfrac{x^2+x+1}{x^2-x+1}\bigg) using~chain~rule

=\dfrac{x^2-x+1}{x^2+x+1}\times\left\{\dfrac{(x^2-x+1)\dfrac{d}{dx}(x^2+x+1)-(x^2+x+1)\dfrac{d}{dx}(x^2-x+1)}{(x^2-x+1)^2}\right\}

=\dfrac{x^2-x+1}{x^2+x+1}\times\left\{\dfrac{(x^2-x+1)(2x+1)-(x^2+x+1)(2x-1)}{(x^2-x+1)^2}\right\}

=\dfrac{x^2-x+1}{x^2+x+1}\times\left\{\dfrac{2x^3-2x^2+2x+x^2-x+1+x^2+x+1-2x^3-2x^2-2x}{(x^2-x+1)^2}\right\}

=\dfrac{x^2-x+1}{x^2+x+1}\times\left\{\dfrac{-4x^2+2x^2+2}{(x^2-x+1)^2}\right\}

=\left\{\dfrac{-4x^2+2x^2+2}{(x^2+x+1)(x^2-x+1)}\right\}

=\dfrac{-2(x^2-1)}{(x^2+x+1)(x^2-x+1)}

=\dfrac{-2(x^2-1)}{(x+1)^2-(x^2)}

=\dfrac{-2(x^2-1)}{x^4+1+2x^2-x^2}

=\dfrac{-2(x^2-1)}{x^4+x^2+1}

So, \dfrac{d}{dx}\left\{log~\bigg(\dfrac{x^2+x+1}{x^2-x+1}\bigg)\right\}=\dfrac{-2(x^2-1)}{x^4+x^2+1}

 

 

Question 33

Differentiate

{tan}^{-1}~(e^x)

Sol :

Let y={tan}^{-1}~(e^x)

Differentiating y with respect to x we get,

\dfrac{dy}{dx}=\dfrac{d}{dx}({tan}^{-1}~(e^x))

=\dfrac{1}{1+(e^x)^2}\times\dfrac{d}{dx}(e^x) using~chain~rule

=\dfrac{e^x}{1+e^{2x}}

So, \dfrac{d}{dx}\left\{{tan}^{-1}~(e^x)\right\}=\dfrac{e^x}{1+e^{2x}}

 

 

Question 34

Differentiate

e^{sin^{-1}~2x}

Sol :

Let y=e^{sin^{-1}~2x}

Differentiating y with respect to x we get,

\dfrac{dy}{dx}=\dfrac{d}{dx}(e^{sin^{-1}~2x})

=(e^{sin^{-1}~2x})\times\dfrac{d}{dx}(sin^{-1}~2x) using~chain~rule

=(e^{sin^{-1}~2x})\times\dfrac{1}{\sqrt{1-(2x)^{2}}}\times\dfrac{d}{dx}(2x)

=\dfrac{2~x~e^{sin^{-1}~2x}}{\sqrt{1-4x^2}}

So, \dfrac{d}{dx}\left\{{tan}^{-1}~(e^x)\right\}=\dfrac{e^x}{1+e^{2x}}

 

Question 35

Differentiate

sin~(2~sin^{-1}~x)

Sol :

Let y=sin~(2~sin^{-1}~x)

Differentiating y with respect to x we get,

\dfrac{dy}{dx}=\dfrac{d}{dx}(sin~(2~sin^{-1}~x))

=cos~(2~sin^{-1}~x)\times\dfrac{d}{dx}(2~sin^{-1}~x)) using~chain~rule

=cos~(2~sin^{-1}~x)\times2\times \dfrac{1}{\sqrt{1-x^2}}

=\dfrac{2~cos~(2~sin^{-1}~x)}{\sqrt{1-x^2}}

So, \dfrac{d}{dx}\left\{sin~(2~sin^{-1}~x)\right\}=\dfrac{e^x}{1+e^{2x}}

 

 

Question 36

Differentiate

e^{tan^{-1}~\sqrt{x}}

Sol :

Let y=e^{tan^{-1}~\sqrt{x}}

Differentiating y with respect to x we get,

\dfrac{dy}{dx}=\dfrac{d}{dx}(e^{tan^{-1}~\sqrt{x}})

=(e^{tan^{-1}~\sqrt{x}})\times\dfrac{d}{dx}({tan^{-1}~\sqrt{x}}) using~chain~rule

=(e^{tan^{-1}~\sqrt{x}})\times\dfrac{1}{1+x}\times\dfrac{d}{dx}(\sqrt{x}) using~chain~rule

=(e^{tan^{-1}~\sqrt{x}})\times\dfrac{1}{1+x}\times\dfrac{1}{2\sqrt{x}}

=\dfrac{e^{tan^{-1}~\sqrt{x}}}{(1+x)~2\sqrt{x}}

So, \dfrac{d}{dx}\left\{e^{tan^{-1}~\sqrt{x}}\right\}=\dfrac{e^{tan^{-1}~\sqrt{x}}}{(1+x)~2\sqrt{x}}

 

 

Question 37

Differentiate

\sqrt{tan^{-1}~\bigg(\dfrac{x}{2}\bigg)}

Sol :

Let y=\sqrt{tan^{-1}~\bigg(\dfrac{x}{2}\bigg)}

Differentiating y with respect to x we get,

\dfrac{dy}{dx}=\dfrac{d}{dx}\bigg\{{tan^{-1}~\bigg(\dfrac{x}{2}\bigg)}~\bigg\}^\frac{1}{2}

=\dfrac{1}{2}\times{\bigg\{tan^{-1}~\bigg(\dfrac{x}{2}\bigg)}\bigg\}^{\frac{1}{2}-1}\times\dfrac{d}{dx}\bigg\{{tan^{-1}~\bigg(\dfrac{x}{2}\bigg)}\bigg\} using~chain~rule

=\dfrac{1}{2}\times{\bigg\{tan^{-1}~\bigg(\dfrac{x}{2}\bigg)}\bigg\}^{\frac{1}{2}-1}\times\dfrac{1}{1+\bigg(\dfrac{x}{2}\bigg)^2}\times \dfrac{1}{2}

=\dfrac{1}{4}\times{\bigg\{tan^{-1}~\bigg(\dfrac{x}{2}\bigg)}\bigg\}^{\frac{-1}{~2}}\times\dfrac{1}{1+\bigg(\dfrac{x}{2}\bigg)^2}

=\dfrac{1}{4}\times{\bigg\{tan^{-1}~\bigg(\dfrac{x}{2}\bigg)}\bigg\}^{\frac{-1}{~2}}\times\dfrac{4}{4+x^2}

={\dfrac{1}{\sqrt{tan^{-1}~\bigg(\dfrac{x}{2}}\bigg)}}\times\dfrac{1}{4+x^2}

={\dfrac{1}{(4+x^2)\sqrt{tan^{-1}~\bigg(\dfrac{x}{2}}\bigg)}}

So, \dfrac{d}{dx}\left\{\sqrt{tan^{-1}~\bigg(\dfrac{x}{2}\bigg)}\right\}={\dfrac{1}{(4+x^2)\sqrt{tan^{-1}~\bigg(\dfrac{x}{2}}\bigg)}}

 

Question 38

Differentiate

log~(tan^{-1}~x)

Sol :

Let y=log~(tan^{-1}~x)

Differentiating y with respect to x we get,

\dfrac{dy}{dx}=\dfrac{d}{dx}\bigg(log~(tan^{-1}~x)\bigg)

=\dfrac{1}{tan^{-1}~x}\times\dfrac{d}{dx}(tan^{-1}~x) using~chain~rule

=\dfrac{1}{tan^{-1}~x}\times\dfrac{1}{1+x^2}

=\dfrac{1}{tan^{-1}~x~(1+x^2)}

So, \dfrac{d}{dx}\left\{log~(tan^{-1}~x)\right\}=\dfrac{1}{tan^{-1}~x~(1+x^2)}

 

 

Question 39

Differentiate

\dfrac{2^x~cos~x}{(x^2+3)^2}

Sol :

Let y=\dfrac{2^x~cos~x}{(x^2+3)^2}

Differentiating y with respect to x we get,

\dfrac{dy}{dx}=\dfrac{d}{dx}\bigg(\dfrac{2^x~cos~x}{(x^2+3)^2}\bigg)

=\left\{\dfrac{(x^2+3)^2\times\dfrac{d}{dx}\times(2^x~cos~x)-(2^x~cos~x)\times\dfrac{d}{dx}\times(x^2+3)^2}{[(x^2+3)^2]^2}\right\} using~product~rule

=\left[\dfrac{(x^2+3)^2\bigg\{(2^x)\times\dfrac{d}{dx}\times(cos~x)+(cos~x)\times\dfrac{d}{dx}\times(2^x)\bigg\}-(2^x~cos~x)\times2(x^2+3)\times\dfrac{d}{dx}(x^2+3)}{[(x^2+3)^2]^2}\right]

=\left[\dfrac{(x^2+3)^2\bigg\{(2^x)\times(-sin~x)+(cos~x)\times(2^x~log_e~2)\bigg\}-(2^x~cos~x)\times 2(x^2+3)\times(2x)}{[(x^2+3)^2]^2}\right]

=2^x(x^2+3)\left[\dfrac{(x^2+3)\bigg\{-sin~x+cos~x~log_e~2\bigg\}-(4x~cos~x)}{(x^2+3)^4}\right] taking common 2^x and (x^2+3)

=\dfrac{2^x}{(x^2+3)^2}\left[cos~x~log_e~2-sin~x-\dfrac{4x~cos~x}{{(x^2+3)}}\right]

So, \dfrac{d}{dx}\left\{\dfrac{2^x~cos~x}{(x^2+3)^2}\right\}=\dfrac{2^x}{(x^2+3)^2}\left[cos~x~log_e~2-sin~x-\dfrac{4x~cos~x}{{(x^2+3)}}\right]

 

 

Question 40

Differentiate

x~sin~2x+5^x+k^k+(tan^2~x)^3

Sol :

Let y=x~sin~2x+5^x+k^k+(tan^2~x)^3

Differentiating y with respect to x we get,

\dfrac{dy}{dx}=\dfrac{d}{dx}\bigg\{x~sin~2x+5^x+k^k+(tan^2~x)^3\bigg\}

=\dfrac{d}{dx}(x~sin~2x)+\dfrac{d}{dx}(5^x)+\dfrac{d}{dx}(k^k)+\dfrac{d}{dx}(tan^2~x)^3

=\dfrac{d}{dx}(x~sin~2x)+\dfrac{d}{dx}(5^x)+\dfrac{d}{dx}(k^k)+\dfrac{d}{dx}(tan~x)^6

=\dfrac{d}{dx}(x~sin~2x)+\dfrac{d}{dx}(5^x)+\dfrac{d}{dx}(k^k)+6(tan~x)^{5-1}\dfrac{d}{dx}(tan~x)

=\bigg\{(x)\dfrac{d}{dx}(sin~2x)+(sin~2x)\dfrac{d}{dx}(x)\bigg\}+(5^x~log_e~5)+0+6(tan~x)^{4}\times(sec^2~x)

=\bigg\{2x~cos~2x+sin~2x\bigg\}+ 5^x~log_e~5+6(tan~x)^{4}\times(sec^2~x)

=2x~cos~2x+sin~2x+ 5^x~log_e~5+6(tan~x)^{4}\times(sec^2~x)

So, \dfrac{d}{dx}\left\{x~sin~2x+5^x+k^k+(tan^2~x)^3\right\}=2x~cos~2x+sin~2x+ 5^x~log_e~5+6(tan~x)^{4}\times(sec^2~x)

 

 

Question 41

Differentiate

log~(3x+2)-x^2~log~(2x-1)

Sol :

Let y=log~(3x+2)-x^2~log~(2x-1)

Differentiating y with respect to x we get,

\dfrac{dy}{dx}=\dfrac{d}{dx}\bigg\{log~(3x+2)-x^2~log~(2x-1)\bigg\}

=\dfrac{1}{3x+2}\times\dfrac{d}{dx}(3x+2)-\dfrac{d}{dx}\bigg\{x^2~log~(2x-1)\bigg\}

=\dfrac{3}{3x+2}-\bigg\{(x^2)\times\dfrac{d}{dx}\times\{log~(2x-1)\}+\{log~(2x-1)\}\times\dfrac{d}{dx}\times(x^2)\bigg\}

=\dfrac{3}{3x+2}-\bigg\{(x^2)\times\dfrac{1}{2x-1}\times 2 +\{log~(2x-1)\}\times 2x\bigg\}

=\dfrac{3}{3x+2}-\bigg\{\dfrac{2~x^2}{2x-1}+2x~log~(2x-1)\bigg\}

=\dfrac{3}{3x+2}-\dfrac{2~x^2}{2x-1}-2x~log~(2x-1)

So, \dfrac{d}{dx}\left\{log~(3x+2)-x^2~log~(2x-1)\right\}=\dfrac{3}{3x+2}-\dfrac{2~x^2}{2x-1}-2x~log~(2x-1)

 

 

Question 42

Differentiate

\dfrac{3x^2~sin~x}{\sqrt{7-x^2}}

Sol :

Let y=\dfrac{3x^2~sin~x}{\sqrt{7-x^2}}

Differentiating y with respect to x we get,

\dfrac{dy}{dx}=\dfrac{d}{dx}\bigg\{\dfrac{3x^2~sin~x}{\sqrt{7-x^2}}\bigg\}

=\left\{\dfrac{(\sqrt{7-x^2})\times\dfrac{d}{dx}(3x^2~sin~x)-(3x^2~sin~x)\times\dfrac{d}{dx}(\sqrt{7-x^2})}{(\sqrt{7-{x^2})}^2}\right\} using~quotient~rule

=\left[\dfrac{(\sqrt{7-x^2})\times\bigg\{(3x^2)\times\dfrac{d}{dx}(sin~x)+(sin~x)\times\dfrac{d}{dx}(3x^2)\bigg\}-(3x^2~sin~x)\times\dfrac{1}{2\sqrt{7-x^2}}\times\dfrac{d}{dx}(7-x^2)}{7-x^2}\right]

=\left[\dfrac{(\sqrt{7-x^2})\times\bigg\{(3x^2)\times(cos~x)+(sin~x)\times(6x)\bigg\}-(3x^2~sin~x)\times\dfrac{1}{2\sqrt{7-x^2}}\times(-2x)}{(7-x^2)}\right]

=\left[\dfrac{(\sqrt{7-x^2})\times\bigg\{3x^2~cos~x+6x~sin~x\bigg\}+\dfrac{6x^3~sin~x}{2\sqrt{7-x^2}}}{(7-x^2)}\right] \dfrac{d}{dx}\sqrt{x}=\dfrac{1}{2\sqrt{x}}

=\dfrac{\sqrt{7-x^2}\times 3x\bigg\{x~cos~x+2~sin~x\bigg\}}{(7-x^2)}+\dfrac{3x^3~sin~x}{\sqrt{7-x^2}~(7-x^2)}

=\dfrac{3x\bigg\{x~cos~x+2~sin~x\bigg\}}{\sqrt{7-x^2}}+\dfrac{3x^3~sin~x}{(7-x^2)^{\frac{3}{2}}}

=\dfrac{3x^2~cos~x+6x~sin~x}{\sqrt{7-x^2}}+\dfrac{3x^3~sin~x}{(7-x^2)^{\frac{3}{2}}}

So, \dfrac{d}{dx}\left\{\dfrac{3x^2~sin~x}{\sqrt{7-x^2}}\right\}=\dfrac{3x^2~cos~x+6x~sin~x}{\sqrt{7-x^2}}+\dfrac{3x^3~sin~x}{(7-x^2)^{\frac{3}{2}}}

 

 

Question 43

Differentiate

sin^2~\{log~(2x+3)\}

Sol :

Let y=sin^2~\{log~(2x+3)\}

Differentiating y with respect to x we get,

\dfrac{dy}{dx}=\dfrac{d}{dx}\bigg\{[sin~\{log~(2x+3)\}]^2\bigg\}

=2~sin~\{log~(2x+3)\}\times\dfrac{d}{dx}[sin~\{log~(2x+3)\}] using~chain~rule

=2~sin~\{log~(2x+3)\}\times cos~\{log~(2x+3)\}\times\dfrac{d}{dx}\{log~(2x+3)\}

=2~sin~\{log~(2x+3)\}\times cos~\{log~(2x+3)\}\times\dfrac{1}{(2x+3)}\times\dfrac{d}{dx}(2x+3)

=sin~2\{log~(2x+3)\} \times\dfrac{2}{(2x+3)} \because 2sin~x~cos~x=sin~2x

So, \dfrac{d}{dx}\left\{sin^2~\{log~(2x+3)\}\right\}=sin~2\{log~(2x+3)\} \times\dfrac{2}{(2x+3)}

 

 

Question 44

Differentiate

e^x~log~sin~2x

Sol :

Let y=e^x~log~sin~2x

Differentiating y with respect to x we get,

\dfrac{dy}{dx}=\dfrac{d}{dx}\bigg\{e^x~log~sin~2x\bigg\}

=(e^x)\times\dfrac{d}{dx}(log~sin~2x)+(log~sin~2x)\times\dfrac{d}{dx}(e^x) using~product~rule

=(e^x)\times\dfrac{1}{sin~2x}\times\dfrac{d}{dx}(sin~2x)+(log~sin~2x)\times(e^x) using~chain~rule

=(e^x)\times\dfrac{1}{sin~2x}\times cos~2x\times\dfrac{d}{dx}(2x)+(log~sin~2x)\times(e^x) using~chain~rule

=e^x~.~2~cot~2x+e^x~.~log~sin~2x

So, \dfrac{d}{dx}\left\{e^x~log~sin~2x\right\}=e^x~.~2~cot~2x+e^x~.~log~sin~2x

 

 

Question 45

Differentiate

\dfrac{\sqrt{x^2+1}+\sqrt{x^2-1}}{\sqrt{x^2+1}-\sqrt{x^2-1}}

Sol :

on rationalising , we get

=\dfrac{\sqrt{x^2+1}+\sqrt{x^2-1}}{\sqrt{x^2+1}-\sqrt{x^2-1}}\times \dfrac{\sqrt{x^2+1}+\sqrt{x^2-1}}{\sqrt{x^2+1}+\sqrt{x^2-1}}

=\dfrac{\left(\sqrt{x^2+1}+\sqrt{x^2-1}\right)^2}{\left(\sqrt{x^2+1}\right)^2-\left(\sqrt{x^2-1}\right)^2}

 

Let y=\dfrac{\sqrt{x^2+1}+\sqrt{x^2-1}}{\sqrt{x^2+1}-\sqrt{x^2-1}}

Differentiating y with respect to x we get,

\dfrac{dy}{dx}=\dfrac{d}{dx}\bigg\{e^x~log~sin~2x\bigg\}

=(e^x)\times\dfrac{d}{dx}(log~sin~2x)+(log~sin~2x)\times\dfrac{d}{dx}(e^x) using~product~rule

=(e^x)\times\dfrac{1}{sin~2x}\times\dfrac{d}{dx}(sin~2x)+(log~sin~2x)\times(e^x) using~chain~rule

 

 

So, \dfrac{d}{dx}\left\{\dfrac{\sqrt{x^2+1}+\sqrt{x^2-1}}{\sqrt{x^2+1}-\sqrt{x^2-1}}\right\}=

 

 

Question 46

Differentiate

log~\left\{x+2+\sqrt{x^2+4x+1}\right\}

Sol :

 

Question 47

Differentiate

(sin^{-1}~x^4)^4

Sol :

 

 

Question 48

Differentiate

sin^{-1}~\bigg(\dfrac{x}{\sqrt{x^2+a^2}}\bigg)

Sol :

 

 

Question 49

Differentiate

\dfrac{e^x~sin~x}{(x^2+2)^3}

Sol :

 

 

Question 50

Differentiate

3~e^{-3x}~log~(1+x)

Sol :

 

 

Question 51

Differentiate

\dfrac{x^2+2}{\sqrt{cos~x}}

Sol :

 

 

Question 52

Differentiate

Sol :

 

 

Question 53

Differentiate

Sol :

 

Question 54

Differentiate

Sol :

 

 

Question 55

Differentiate

Sol :

 

 

Question 56

Differentiate

Sol :

 

 

Question 57

Differentiate

Sol :

 

 

Question 58

Differentiate

Sol :

 

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