Class 12 ncert solution math exercise 5.7 differentiation

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   Exercise 5.7 ( Differentiation)

Find the second order derivatives of the functions given in Exercises 1 to 10.(Class 12 ncert solution math exercise 5.7 differentiation)

Question:1 x^2 + 3x + 2

Solution: Let y = x^2 + 3x + 2

Then, differentiate with respect to x

\frac{dy}{dx} = \frac{d}{dx}(x^2+3x+2)

= 2x +3

Again differentiate w.r. to x

\frac{d^2y}{dx^2} = \frac{d}{dx}(2x + 3)

= 2

Question 2: x^{20}

Solution: Let y = x^{20}

Diffrentiate with respect to x

\frac{dy}{dx} = \frac{d}{dx}x^{20}

= {20} x^{19}

Again d.w.r. to x

\frac{d^2y}{dx^2} = 20\times19 x^{18} = 380x^{18}

Question 3: x.\cos x

Solution: Let y = x.\cos x

Differentiate .w.r.to x

\frac{dy}{dx}=\frac{d}{dx}x.\cos x

= x\frac{d}{dx}\cos x + \cos x\frac{d}{dx} x

= x (-\sin x) + \cos x.1

= -x\sin x + \cos x

Again differentiate with respect to x

\frac{d^2y}{dx^2} = \frac{d}{dx}(-x\sin x + \cos x)

= \frac{d}{dx}(-x.\sin x)+\frac{d}{dx}\cos x

=-(x\frac{d}{dx}\sin x+\sin x\frac{d}{dx}x) -\sin x

= -(x.\cos x + \sin x.1)-\sin x

=-x.\cos x -\sin x -\sin x

= -x\cos x - 2\sin x

Question 4: \log x

Solution: Let y=\log x

Differentiate with respect to x

\frac{dy}{dx}=\frac{d}{dx}\log x

=\frac{1}{x}

Again differentiate with respect to x

\frac{d^2y}{dx^2}=\frac{d}{dx}\frac{1}{x}

= \frac{d}{dx}x^{-1}
= (-1x^{-2})
= \frac{-1}{x^2}

Question 5: x^3 \log x

Solution: Let y=x^3\log x

Differentiate with respect to x

\frac{dy}{dx}=\frac{d}{dx}x^3\log x

=x^3\frac{d}{dx}\log x+\log x\frac{d}{dx}x^3

=x^3\frac{1}{x}+\log x.3x^2

= x^2 +3x^2\log x

Again differentiate with respect to x

\frac{d^2y}{dx^2}=\frac{d}{dx}x^2+\frac{d}{dx}3x^2\log x

= 2x +3(x^2\frac{d}{dx}\log x+\log x\frac{d}{dx}x^2)

= 2x+ 3(x^2.\frac{1}{x}+\log x.2x)

= 2x+3x+6x\log x

=5x+6x\log x

Question 6: e^x \sin 5x

Solution: Let y=e^x \sin 5x

Differentiate with respect to x

\frac{dy}{dx} = \frac{d}{dx}e^x \sin 5x

=e^x\frac{d}{dx}\sin 5x+\sin 5x\frac{d}{dx}e^x

= e^x\cos 5x.5+\sin 5x.e^x

= 5e^x\cos 5x+e^x\sin 5x

Again differentiate with respect to x

\frac{d^2y}{dx^2}=\frac{d}{dx}(5e^x\cos 5x+e^x\sin 5x)

= 5(e^x\frac{d}{dx}\cos 5x+\cos 5x\frac{d}{dx}e^x)+(e^x\frac{d}{dx}\sin 5x+\sin 5x\frac{d}{dx}e^x)

=5(-e^x\sin 5x.5 +\cos 5x.e^x)+(e^x.\cos 5x.5+\sin 5x.e^x)

=-25e^x\sin 5x+5e^x\cos 5x+5e^x\cos5x+e^x\sin 5x

=-24e^x\sin 5x+10e^x\cos 5x

= 2e^x(5\cos 5x -12\sin 5x)

Question 7: e^{6x} cos 3x

Solution: Let y = e^{6x}\cos 3x

\frac{dy}{dx} =\frac{d}{dx}e^{6x}\cos 3x

= e^{6x}\frac{d}{dx}\cos 3x+\cos 3x\frac{d}{dx}e^{6x}
= e^{6x}(-\sin 3x.3)+\cos 3x(6e^{6x})
=3e^{6x}(2\cos 3x-\sin 3x)

Again differentiate with respect to x

\frac{d^2y}{dx^2}=\frac{d}{dx}3e^{6x}(2\cos 3x-\sin 3x)

= 3e^{6x}\frac{d}{dx}(2\cos 3x-\sin 3x)+3(2\cos 3x-\sin 3x)\frac{d}{dx}e^{6x}

=3e^{6x}(-2\sin 3x.3-\cos 3x.3)+3(2\cos 3x-\sin 3x)6e^{6x}

= e^{6x}(-18\sin 3x-9\cos 3x+36\cos 3x - 18\sin 3x)
=e^{6x}(-36\sin 3x+27\cos 3x)
=9e^{6x}(3\cos 3x - 4\sin 3x)

Question 8: \tan^{-1}x

Solution: Let y = \tan^{-1}x

Differentiate with respect to x

\frac{dy}{dx}= \frac{d}{dx}\tan^{-1}x

=\frac{1}{1+x^2}

Again differentiate with respect to x

\frac{d^2y}{dx^2} =\frac{d}{dx}\frac{1}{1+x^2}

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

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

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

= \frac{-2x}{1+x^2}

Question 9: log (log x)

Solution: Let y=log (log x)

Differentiate with respect to x

\frac{dy}{dx} = \frac{d}{dx}log (log x)

=\frac{1}{\log x}\frac{d}{dx}\log x

= \frac{1}{\log x}.\frac{1}{x}

= \frac{1}{x.\log x}

Again differentiate with respect to x

\therefore \frac{d^2y}{dx^2}=\frac{d}{dx}(x.\log x)^{-1}

= -1(x.\log x)^{-2}\frac{d}{dx}x.\log x

= \frac{-1}{(x\log x)^2}(x\frac{d}{dx}\log x + \log x\frac{d}{dx}x)

= \frac{-1}{(x.\log x)^2}(x\frac{1}{x}+\log x.1)

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

Question10: \sin(\log x)

Solution: Let y= \sin(\log x)

Differentiate with respect to x

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

= \cos(\log x)\frac{d}{dx}\log x

= \cos(\log x)\frac{1}{x}

=\frac{\cos(\log x)}{x}

Again differentiate with respect to x

\therefore \frac{d^2y}{dx^2}=\frac{d}{dx}\frac{\cos(\log x)}{x}

=\frac{x\frac{d}{dx}\cos(\log x)-\cos(\log x)\frac{d}{dx}x}{x^2}

= \frac{-x\sin(\log x)\frac{1}{x}-\cos(\log x).1}{x^2}

= \frac{-[\sin(\log x)+\cos(\log x)]}{x^2}

Question 11: If y = 5 \cos x - 3 \sin x, prove that \frac{d^2y}{dx^2}+y=0

Solution: Let y = 5 \cos x - 3 \sin x

Differentiate with respect to x

\frac{dy}{dx}=5\frac{d}{dx}\cos x-3\frac{d}{dx}\sin x

= -5\sin x-3\cos x

Again differentiate with respect to x

\frac{d^2y}{dx^2}=\frac{d}{dx}( -5\sin x-3\cos x)

= -5\cos x +3\sin x

\Rightarrow \frac{d^2y}{dx^2} =-(5\cos 3x-3\sin 3x)

\Rightarrow \frac{d^2y}{dx^2}=-y

\Rightarrow \frac{d^2y}{dx^2}+y=0

Question 12: If y =\cos^{-1}x, Find \frac{d^2y}{dx^2} in term of y alone.

Solution: Since, y=\cos^{-1}x
\cos y = x

Differentiate with respect to x

\frac{d}{dx}\cos y=\frac{d}{dx}x

-\sin y\frac{dy}{dx}= 1
\Rightarrow \frac{dy}{dx} = \frac{-1}{\sin y}

\Rightarrow \frac{dy}{dx}=-\operatorname{cosec y}

Again differentiate with respect to x

\therefore \frac{d^2y}{d^2x}=-\frac{d}{dx}\operatorname{cosec y}

=-(-\operatorname{cosec y}\cot y)\frac{dy}{dx}

= +\operatorname{cosec y}\cot y\operatorname{(-cosec y)}

=-\operatorname{cosec^2 y}\cot y

Question 13: If y = 3 \cos (\log x) + 4 \sin (\log x), show that x^2 y_2 + xy_1 + y = 0

Solution: Since, y = 3 \cos (\log x) + 4 \sin (\log x),

Differentiate with respect to x

\frac{dy}{dx}=\frac{d}{dx}[3 \cos (\log x) + 4 \sin (\log x)]

=3\frac{d}{dx}\cos(\log x)+4\frac{d}{dx}\sin(\log x)

=-3\sin(\log x)\frac{d}{dx}\log x+4\cos(\log x)\frac{d}{dx}\log x

\Rightarrow \frac{dy}{dx}= -3\sin(\log x)\frac{1}{x}+4\cos(\log x)\frac{1}{x}

Multiply by x both side

x\frac{dy}{dx}=-3\sin(\log x)+4\cos(\log x)

Again differentiate with respect to x

x\frac{d^2y}{dx^2}+\frac{dy}{dx}.1=-3\frac{d}{dx}\sin(\log x)+4\frac{d}{dx}\cos(\log x)

\Rightarrow x\frac{d^2y}{dx^2}+\frac{dy}{dx}=-3\cos(\log x).\frac{1}{x}-4\sin(\log x).\frac{1}{x}

Agian multiply by x

\Rightarrow x^2\frac{d^2y}{dx^2}+x\frac{dy}{dx}=-[3\cos(\log x)+4\sin(\log x)]

\Rightarrow x^2\frac{d^2y}{dx^2}+x\frac{dy}{dx}=-y

\Rightarrow x^2\frac{d^2y}{dx^2}+x\frac{dy}{dx}+y=0

\Rightarrow y_2 + xy_1 + y = 0

Hence proved

Question 14: If y = Ae^{mx} + Be^{nx}, show that \frac{d^2y}{dx^2}-(m+n)\frac{dy}{dx}+mny=0

Solution: Since, y = Ae^{mx} + Be^{nx},

Differentiate with respect to x

\frac{dy}{dx}=A\frac{d}{dx}e^{mx}+B\frac{d}{dx}e^{nx}

=Ae^{mx}.m+e^{nx}.n

\Rightarrow \frac{dy}{dx}= mAe^{mx}+nBe^{nx}

Again differentiate with respect x

\frac{d^2y}{dx^2}=mA\frac{d}{dx}e^{mx}+nB\frac{d}{dx}e^{nx}

=mAe^{mx}.m+nBe^{nx}.n

\Rightarrow \frac{d2y}{dx^2}=m^2Ae^{mx}+n^2Be^{nx}

L.H.S. \frac{d^2y}{dx^2}-(m+n)\frac{dy}{dx}+mny

Putting the values

(m^2Ae^{mx}+n^2Be^{nx})-(m+n)(mAe^{mx}+nBe^{nx})+mn(Ae^{mx} + Be^{nx})

\Rightarrow m^2Ae^{mx}+n^2Be^{nx}-(m^2Ae^{mx}+mnBe^{nx}+nmAe^{mx}+n^2Be^{nx})+mnAe^{mx}+mnBe^{nx}

\Rightarrow m^2Ae^{mx}+n^2Be^{nx}-m^2Ae^{mx}-mnBe^{nx}-nmAe^{mx}-n^2Be^{nx}+mnAe^{mx}+mnBe^{nx}

= 0
Hence proved

Question 15: If y = 500e^{7x} + 600e^{- 7x}, show that \frac{d^2y}{dx^2}=49y

Solution: Given, y = 500e^{7x} + 600e^{- 7x}

Differentiate with respect to x

\frac{dy}{dx}= 500\frac{d}{dx}e^{7x}+600\frac{d}{dx}e^{-7x}

= 500\times 7e^{7x}+600\times (-7)e^{-7x}

Again differentiate with respect to x

\frac{d^2y}{dx^2}=500\times -7\times -7e^{-7x}+600\times -7 \times -7e^{-7x}

= 500\times 49 e^{7x}+600\times 49e^{-7x}

=49(500e^{7x}+600e^{-7x})

\frac{d^2y}{dx^2}=49y

Hence proved,

Question 16: If e^y(x + 1) = 1, show that \frac{d^2y}{dx^2}=(\frac{dy}{dx})^2

Solution: Given,e^y(x + 1) = 1

e^y=\frac{1}{x+1} ---(1)

Differentiate with respect to x

\frac{d}{dx}e^y=\frac{d}{dx}(x+1)^{-1}

\Rightarrow e^y\frac{dy}{dx}=-1(x+1)^{-2}.1

\Rightarrow \frac{dy}{dx} =\frac{-1}{e^y(x+1)^2}

\Rightarrow\frac{dy}{dx} = \frac{-1}{\frac{1}{(x+1)}(x+1)^2}

\Rightarrow \frac{dy}{dx} = \frac{-1}{(x+1)}

Again differentiate with respect to x

\frac{d^y}{dx^2}=-\frac{d}{dx}(x+1)^{-1}

\Rightarrow \frac{d^2y}{dx^2} = +(x+1)^{-2}

\Rightarrow \frac{d^2y}{dx^2}= \frac{1}{(x+1)^2}

L.H.S. \frac{d^2y}{dx^2}= \frac{1}{(x+1)^2}

R.H.S. (\frac{dy}{dx})^2=(\frac{-1}{(x+1)})^2

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

Hence, L.H.S.=R.H.S.

Proved

Question 17: If y = (tan^{-1} x)^2, show that (x^2 + 1)^2 y_2 + 2x (x^2 + 1) y_1 = 2

Solution: Given, y = (tan^{-1} x)^2

Differentiate with respect to x

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

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

Multiply by (1+x^2) both side

(1+x^2)\frac{dy}{dx}=2\tan^{-1}x

Again differentiate with respect to x

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

\Rightarrow (1+x^2)\frac{d^2y}{dx^2}+\frac{dy}{dx}.2x = 2\frac{1}{1+x^2}

Again multiply by (1+x^2) both side

\Rightarrow (1+x^2)^2\frac{d^2y}{dx^2}+2x(1+x^2)\frac{dy}{dx}=2

\Rightarrow (1+x^2)^2y_2+2x(1+x^2)y_1=2

Hence proved.

Ncert Solution Chapter 5 : Differention

Ncert Solution solution exercise 5.1

Ncert Solution solution exercise 5.2

Ncert Solution solution exercise 5.3

Ncert Solution solution exercise 5.4

Ncert Solution solution exercise 5.5

Ncert Solution solution exercise 5.6

Ncert Solution solution exercise 5.7

Ncert Solution solution Miscellaneous exercise

 

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