Continuity & Differentiability NCERT Solutions

NCERT Exercise 5.7

Find the second order derivatives of the functions given in exercises 1 to 10

Question: (1) `x^2 + 3x+ 2`

Solution:

Let `y= x^2+3x+2`

Differentiating both sides w.r.t. `x`, we have

`(dy)/(dx) = d/ dx(x^2+3x+2)`

`=(2x+3)`

Again differentiating both sides w.r.t. `x` , we have

`d/(dx) ((dy)/(dx))= d/(dx)(2x+3)`

`(d^2y)/(dx^2)=2`

Question: (2) `x ^20`

Solution:

Let `y= x^20`

Differentiating both sides w.r.t. `x`, we have

` (dy)/(dx)= d/(dx)(x^20)`

`= (20 x^19)`

Again differentiating both sides w.r.t. `x` , we have

`d/(dx) ((dy)/(dx))= d/(dx)(20x^19)`

`= 20 xx 19^18=380x^18`

Question: (3) `x.cos x`

Solution:

Let `y=x. cos x`

Differentiating both sides w.r.t. `x`, we have

` (dy)/(dx)= d/(dx)(x.cos x)`

`=x.d/(dx)(cos x)+cs x d/(dx)(x)`

Again differentiating both sides w.r.t. `x` , we have

`d/(dx) ((dy)/(dx))= d/(dx)[cos x- xsin x]`

`(d^2y)/(dx^2)`

`=-sin x-(x d/dx sin x+sin d/dx (x))`

`=-sin x- (x cos x+sin x)`

`=-sin x - x cos x- sin x`

`=-2sin x- cos x`

`=-(2sin x+ xcos x)`

Question: (4) `log x`

Solution:

Let `y=log x`

Differentiating both sides w.r.t. `x`, we have

` (dy)/(dx)= d/(dx)(log x)=1/x`

Again differentiating both sides w.r.t. `x` , we have

`d/(dx) ((dy)/(dx))= d/(dx)(1/x)`

`(d^2y)/(dx^2)`

`=- 1/x^2`

Question: (5)`x^2 log x`

Solution:

Let `y=x^2 log x`

Differentiating both sides w.r.t. `x`, we have

` (dy)/(dx)= d/(dx)(x^2 log x)`

`x^3.d/dx(log x)+ log x. d/dx(x^3)`

`x^3 .1/x + log x. 3x^2`

`= x^2+ 3x^2 log x`

`x^2 (1+3 log x)`

Again differentiating both sides w.r.t. `x` , we have

`d/(dx) ((dy)/(dx))= d/(dx)[x^2(1+3 log x)]`

`(d^2y)/(dx^2)`

`=x^2. d/dx(1+3 log x)` `+(1+3 log x).d/dx(x^2)`

`=x^2 (3/x)+(1+3 log x).2x`

` 3x+2x + 6xlogx`

`=5x+6xlog x`

Question: (6) `e^x sin 5x`

`Let y= e^x sin 5x`

Differentiating both sides w.r.t. `x`, we have

` (dy)/(dx)= d/(dx)(e^x sin 5x)`

`=e^x .d/dx (sin 5x)` `+ sin 5x . d/dx (e^x)`

` 5e^x cos 5x + e^x. sin 5x`

` e^x(5 cos 5x + sin 5x)`

Again differentiating both sides w.r.t. `x` , we have

`d/(dx) ((dy)/(dx))`

`= d/(dx)[e^x(5 cos 5x +sin 5x)]`

`(d^2y)/(dx^2)`

`e^x. d/dx (5cos 5x+ sin 5x)`

`+ (5cos 5x + sin 5x + si 5x). d/(dx)(e^x)`

`=e^(-5 sin 5x. d/(dx)(5x)+ cos 5x .d/dx(5x)`

`+(5cos 5x + sin 5x).(e^x)`

`e^x(-25sin5x+5cos5x)+(5 cos 5x+ sin5x)e^x`

`=e^x[-25 sin 5x+5 Cos 5x + 5 cos5x+ sin 5x]`

`= 2e^x[5 cos 5x -12 sin5x]`

Question: (8) `tan ^-1x`

Solution:

Let `y=tan^-1 x`

Differentiating both sides w.r.t. `x`, we have

` (dy)/(dx)= d/(dx)(tan ^-1 x)`

`= 1/ (1+x^2)`

Again differentiating both sides w.r.t. `x` , we have

`d/(dx) ((dy)/(dx))= d/(dx)(1/(1+x^2))`

`(d^2y)/(dx^2)`

`= (-2x)/((1+x^2)^2)`

Question: (9) `log(log x)`

Solution:

Let `y= log (log x)`

Differentiating both sides w.r.t. `x`, we have

` (dy)/(dx)= d/(dx)[log(log x)]`

`= 1/(log x) .d/(dx) (log x)`

`=1/(x log x)`

Again differentiating both sides w.r.t. `x` , we have

`d/(dx) ((dy)/(dx))= d/(dx)(1/(x log x))`

`(d^2y)/(dx^2)`

`= (xlog x. d/dx (1) - 1. d/(dx) (x log x))/((x log x)^2)`

`=(-x. d/dx (log x)- log x d/(dx)(x))/((x log x)^2)`

`=(-x xx 1/x -log x.1)/((x log x)^2)`

`= (-(1+ log x))/ (x log x)^2`

Question: (10) `sin (log x)`

Solution:

Let `y= sin (log x)`

Differentiating both sides w.r.t. `x`, we have

` (dy)/(dx)= d/(dx)[sin(log x)]`

`= cos(log x) .d/(dx) (log x)=1/x cos (log x)`

Again differentiating both sides w.r.t. `x` , we have

`d/(dx) ((dy)/(dx))= d/(dx)[x^-1 cos ( log x)]`

`=x^-1.d/(dx)[cos(log x)]+cos (log x).d/dx (x^-1)`

`=1/x. - sin (log x) .d/(dx)(log x)+ cos (log x).(-1)/x^2`

`=-sin(log x)/x^2 -(cos x(log x))/x^2`

`=([-sin(log x)+cos(log x)])/x^2`

Question: (11)  If  `y= 5 cos x-3 sin x`, prove that `(d^2y)/dx^2+y=0`

Solution:

Here `y= 5 cos x- 3 sin x`

Differentiating both sides w.r.t. `x`, we have

` (dy)/(dx)= d/(dx)(5 cos x- 3 sin x)`

`-5 sin x -3 cos x`

Again differentiating both sides w.r.t. `x` , we have

`d/(dx) ((dy)/(dx))= d/(dx)(-5 sin x- 3 cos x)`

`(d^2y)/dx^2 `

`= -5 cos x+ 3 sin x`

`= -(5cos x- 3 sin x)`

`(d^2y)/dx^2 `

`=-y=> (d^2y)/dx^2 +y=0`

Question: (12) If  `y= cos_1 x`, find `(d^2y)/dx^2` in terms of `y` alone.

Solution:

Here `y= cos^-1x`

Differentiating both sides w.r.t. `x`, we have

` (dy)/(dx)= d/(dx)(cos^-1x)`

=`(-1)/(sqrt(1-x^2)`

`= -(1-x^2)^(-1/2)`

Again differentiating both sides w.r.t. `x` , we have

`d/(dx) ((dy)/(dx))= d/(dx)[-(1-x^2)^(-1/2)]`

`(d^2y)/dx^2 `

`=-(-1/2)(1-x^2)^(-3/2).d/dx(1-x^2)`

`=1/(2(1-x^2)^(3//2). -2x`

`= (-1)/(1-x^2^(3//2)`

Now `y=cos^-1x => x= cos y`

`:. (d^2y)/dx^2`

`=(-cos y)/(1-cos^2y)^3/2`

`=(-cos y)/(sin ^2y)^(3//2`

`=(-cos y)/(sin^3 y)`

`= (-cos y)/(sin y. sin^2 y)`

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

Solution:

Here ` y=3 cos (log x)+4 sin (log x)`

Differentiating both sides w.r.t. `x`, we have

` (dy)/(dx)= d/(dx)[3cos(log x)+4 sin (log x)]`

` =-3 sin (log x).d/(dx) (log x)+4 cos(log x).d/dx (log x)`

`=-3 sin (log x)/x + 4 cos (log x)/x`

`:. x=dy/dx = -3 sin (log x)+ 4 cos (log x)`

Again differentiating both sides w.r.t. `x` , we have

`d/(dx) ((dy)/(dx))= d/(dx)[-sin(log x) + 4 cos (log x)]`

`x.d/dx (dy/dx)+ dy/dx .d/dx(x)`

`=-3 cos(log x). d/dx (log x)-4 sin (log x). d/dx(log x)`

`(d^2y)/dx^2+ dy/dx `

`=-3cos (log x)/x - (4 sin(log x)/x`

`x^2 (d^2y)/dx^2+x dy/dx`

`=-[-3 cos (log x)+ 4 sin (log x)]`

`x^2(d^2y)/dx^2 +x dy/dx =-y`

`x^2 (d^2y)/dx^2 +x dy/dx + y=0`

`x^2y_2+ xy_1 + y =0`

Question: (15) If  `y =500e^7x + 600e^-7x` ,show that `(d^2y)/(dx^2)=49y`.

Solution:

Here `Y= 500e^7x+600e^-7x`

Differentiating both sides w.r.t. `x`, we have

` (dy)/(dx)= d/(dx)[500e^(7x)+600e^-7x]`

`500xxe^7x. d/(dx)(7x)+ 600 e^-(7x).d/(dx) (-7x)`

`=3500e^7x- 4200e^-7x`

Again differentiating both sides w.r.t. `x` , we have

`d/(dx) ((dy)/(dx))`

`=d/dx [3500e^7x- 4200e^-7x]`

`(d^2y)/dx^2 `

`= 3500e^7x d/dx(7x)-4200e^-7x d/dx(-7x)`

`=24500e^(7x)-29400e^(7x)`

`=49 (500e^(7x)-600e^(7x)=49y`

Question: (16) If  `e^y(x+1)=1`, show that `(d^2y)/dx^2 =(dy/dx)^2`

Solution:

Differentiating both sides w.r.t. `x`, we have

` d/(dx)[e^y (x+1)]= d/(dx)(1)`

`e^y. d/dx (x+1)+(x+1).d/(dx) (e^y)=0`

`e^y.1(x+1).e^y. dy/ dx =0`

`e^y+ dy/dx =0`

`dy/dx =0`

`dy/dx=-e^y`

Again differentiating both sides w.r.t. `x` , we have

`d/(dx) ((dy)/(dx))`

`d/dx (-e^y)`

`(d^2y)/dx^2`

`=-e^y .dy/dx `

`(d^2y)/dx^2=dy/dx. dy/dx`

`(d^2y)/dx^2 =(dy/ dx)^2`

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

Solution:

Here `y= (tan ^-1 x)^2`

Differentiating both sides w.r.t. `x`, we have

`dy/dx = d/dx [(tan^-1 x)^2]`

`=2 tan^-1x .d/dx (tan ^-1 x)`

`= 2tan ^-1x. 1/(1+x^2)`

`:. (1+ x^2)dy/dx`

`2 tan^-1 x`

Again differentiating both sides w.r.t. `x` , we have

`d/(dx) [(1+x^2) .dy/dx]`

` d/dx [2 tan ^-1 x]`

`(1+x^2).d/dx (dy/dx)+ dy/dx. d/dx(1+x^2)`

`2/(1+x^2)`

`(1+x^2).(d^2y)/(dx^2)+2xdy/dx`

`= 2/ (1+x^2)`

`(1+x^2)^2 (d^2y)/(dx^2)` `+2x(1+x^2)dy/dx =2`

`:. (1+x^2)^2y_2 + 2x (1+x^2)y_1 =2`

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