Class 12 NCERT Solutions- Mathematics Part I – Chapter 5 Continuity And Differentiability – Exercise 5.7

Last Updated : 18 Mar, 2021

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

Question 1. x²+ 3x + 2

Solution:

Here, y = x²+ 3x + 2

First derivative,

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

= 2x+ 3

Second derivative,

$\frac{d^2y}{dx^2} = \frac{d}{dx}(\frac{dy}{dx})$

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

= 2

Question 2. x²⁰

Solution:

Here, y = x²⁰

First derivative,

$\frac{dy}{dx} = \frac{d(x^{20})}{dx}$

= 20x^20-1

= 20x¹⁹

Second derivative,

$\frac{d^2y}{dx^2} = \frac{d}{dx}(\frac{dy}{dx})$

= $\frac{d(20x^{19})}{dx}$

= 20(19x^19-1)

= 380x¹⁸

Question 3. x . cos x

Solution:

Here, y = x . cos x

First derivative,

$\frac{dy}{dx} = \frac{d(x . cos x)}{dx}$

Using product rule

= x $\frac{d(cos x)}{dx}$ + cos x $\frac{d(x)}{dx}$

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

= – x sin x+ cos x

Second derivative,

$\frac{d^2y}{dx^2} = \frac{d}{dx}(\frac{dy}{dx})$

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

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

Using product rule,

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

= -x (cos x) + sin x (-1) – sin x

= – ( x cos x + 2 sin x)

Question 4. log x

Solution:

Here, y = log x

First derivative,

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

= 1/x

Second derivative,

$\frac{d^2y}{dx^2} = \frac{d}{dx}(\frac{dy}{dx})$

= $\frac{d(1/x)}{dx}$

Using division rule,

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

= $\frac{x (0) - 1(1)}{x^2}$

= $\mathbf {\frac{- 1}{x^2}}$

Question 5. x³log x

Solution:

Here, y = x³ . log x

First derivative,

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

Using product rule

= x³ $\frac{d(log (x))}{dx}$ + log x $\frac{d(x^3)}{dx}$

= x³ ( $\frac{1}{x}$ ) + log x (3x²)

= x² + 3x² log x

Second derivative,

$\frac{d^2y}{dx^2} = \frac{d}{dx}(\frac{dy}{dx})$

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

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

Using product rule,

= 2x + 3 (x² $\frac{d(log (x))}{dx}$ – log x $\frac{d(x^2)}{dx}$ )

= 2x + 3 (x² $\frac{1}{x}$ – log x (2x))

= 2x + 3 (x – 2x . log x)

= 2x + 3x – 6x . log x

= x(5 – 6 log x)

Question 6. e^xsin 5x

Solution:

Here, y = e^x sin 5x

First derivative,

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

Using product rule

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

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

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

Second derivative,

$\frac{d^2y}{dx^2} = \frac{d}{dx}(\frac{dy}{dx})$

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

Using product rule,

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

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

= e^x (- 25 sin 5x + 5cos 5x) + (5 cos(5x)+ sin 5x) (e^x)

= e^x (- 25 sin 5x + 5cos 5x + 5 cos(5x)+ sin 5x)

= e^x (10 cos 5x – 24 sin 5x)

Question 7. e^6x cos 3x

Solution:

Here, y = e^6x cos 3x

First derivative,

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

Using product rule

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

= e^6x (- 3 sin(3x))+ cos 3x (6e^6x)

= e^6x (6 cos(3x) – 3 sin (3x))

Second derivative,

$\frac{d^2y}{dx^2} = \frac{d}{dx}(\frac{dy}{dx})$

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

Using product rule,

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

= e^6x (6 (3 (- sin(3x)) – 3 (3 cos 3x)) + (6 cos(3x) – 3 sin (3x)) (6e^6x)

= e^6x (- 18sin(3x) – 9 cos 3x) + (36 cos(3x) – 18 sin (3x)) (e^6x)

= e^6x (27 cos(3x) – 36 sin (3x))

= 9e^6x (3 cos(3x) – 4 sin (3x))

Question 8. tan^–1 x

Solution:

Here, y = tan^–1 x

First derivative,

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

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

Second derivative,

$\frac{d^2y}{dx^2} = \frac{d}{dx}(\frac{dy}{dx})$

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

Using division rule,

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

= $\frac{(x^2+1) . (0) - (2x)}{(x^2+1)^2}$

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

Question 9. log (log x)

Solution:

Here, y = log (log x)

First derivative,

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

= $\frac{1}{log x} \frac{d(log x)}{dx}$

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

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

Second derivative,

$\frac{d^2y}{dx^2} = \frac{d}{dx}(\frac{dy}{dx})$

= $\frac{d}{dx}(\frac{1}{x log x})$

Using division rule,

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

Using product rule,

= $\frac{(x log x) . (0) - (x \frac{d(log x)}{dx} + log x \frac{d(x)}{dx})}{(x log x)^2}$

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

= – $\frac{(1 + log x)}{(x log x)^2}$

= – $\mathbf{\frac{1 + log x}{(x log x)^2}}$

Question 10. sin (log x)

Solution:

Here, y = sin (log x)

First derivative,

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

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

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

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

Second derivative,

$\frac{d^2y}{dx^2} = \frac{d}{dx}(\frac{dy}{dx})$

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

Using division rule,

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

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

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

= $\mathbf{\frac{- sin(log (x)) - cos (log (x))}{x ^2}}$

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

Solution:

Here, y = 5 cos x – 3 sin x

First derivative,

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

= 5 (- sin x) – 3 (cos x)

= – 5 sin(x) – 3 cos(x)

Second derivative,

$\frac{d^2y}{dx^2} = \frac{d}{dx}(\frac{dy}{dx})$

= $\frac{d(- 5 sin(x) - 3 cos(x))}{dx}$

= $\frac{d(- 5 sin(x))}{dx} - \frac{d(3 cos(x))}{dx}$

= -5 (cos(x)) – 3 (- sin(x))

= -(5 cos(x) – 3 sin(x))

= -y

According to the given condition,

$\frac{d^2y}{dx^2}$ + y = -y + y

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

Hence Proved!!

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

Solution:

Here, y = cos-1 x

x = cos y

First derivative,

$\frac{dx}{dy} = \frac{d(cos y)}{dy}$

= – sin y

$\frac{dy}{dx} = \frac{-1}{sin (y)}$

= – cosec (y)

Second derivative,

$\frac{d^2y}{dx^2} = \frac{d}{dx}(\frac{dy}{dx})$

= $\frac{d(- cosec (y))}{dx}$

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

= – (-cosec(y) cot (y)) (-cosec(y))

= -cosec²(y) cot (y)

Hence we get

$\mathbf{\frac{d^2y}{dx^2}}$ = -cosec²(y) cot (y)

Question 13. If y = 3 cos (log x) + 4 sin (log x), show that x² y₂+ xy₁+ y = 0

Solution:

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

First derivative,

y₁ = $\frac{dy}{dx} = \frac{d(3 cos (log x) + 4 sin (log x))}{dx}$

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

= $\frac{1}{x}$ (4 cos (log(x)-3 sin (log x))

Second derivative,

y₂ = $\frac{d^2y}{dx^2} = \frac{d}{dx}(\frac{dy}{dx})$

= $\frac{d}{dx} (\frac{1}{x} (4 cos (log(x)-3 sin (log x)))$

Using product rule.

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

= $\frac{1}{x}$ (4(-sin(log(x))) $\frac{1}{x}$ – 3 (cos(log(x))) $\frac{1}{x}$ ) + (4 cos (log(x)-3 sin (log x)) ( $\frac{1}{x^2}$ )

= $\frac{1}{x}$ (-4sin(log(x)) $\frac{1}{x}$ – 3 cos(log(x)) $\frac{1}{x}$ ) – (4 cos (log(x) + 3 sin (log x)) ( $\frac{1}{x^2}$ )

= \frac{-1}{x^2} [-7 cos(log(x) – sin (log x)]

According to the given conditions,

xy₁ = x( $\frac{1}{x}$ (4 cos (log(x)-3 sin (log x)))

xy₁ = -3 sin (log x)+ 4 cos (log(x))

x² y₂ = x² $(\frac{1}{x^2} [-7 cos(log(x) - sin (log x)])$

x² y₂ =[-7 cos(log(x) – sin (log x)]

Now, rearranging

xy₁ + x² y₂ + y = -3 sin (log x)+ 4 cos (log(x)) + cos(log(x)) -7 cos(log(x) – sin (log x) + 4 sin (log x)

Hence we get

xy₁ + x² y₂ + y = 0

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

Solution:

Here, y = Ae^mx + Be^nx

First derivative,

$\frac{dy}{dx} = \frac{d(Ae^{mx} + Be^{nx})}{dx}$

= mAe^mx + nBe^nx

Second derivative,

$\frac{d^2y}{dx^2} = \frac{d}{dx}(\frac{dy}{dx})$

= $\frac{d}{dx}(mAe^{mx} + nBe^{nx})$

= m²Ae^mx + n²Be^nx

According to the given conditions,

$\frac{d^2y}{dx^2}$ – (m+n) $\frac{dy}{dx}$ + mny, we get

LHS = m²Ae^mx + n²Be^nx – (m+n)(mAe^mx + nBe^nx) + mny

= m²Ae^mx + n²Be^nx – (m²Ae^mx + mnAe^mx + mnBe^nx + n²Be^nx) + mny

= -(mnAe^mx + mnBe^nx) + mny

= -mny + mny

= 0

Hence we get

$\mathbf{\frac{d^2y}{dx^2} - (m+n) \frac{dy}{dx}}$ + mny = 0

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

Solution:

Here, y = 500e^7x+ 600e^{– 7x}

First derivative,

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

= 500e^7x . (7)+ 600e^{– 7x} (-7)

= 7(500e^7x – 600e^{– 7x})

Second derivative,

$\frac{d^2y}{dx^2} = \frac{d}{dx}(\frac{dy}{dx})$

= $\frac{d}{dx}(7(500e^{7x} - 600e^{- 7x}))$

= 7[500e^7x . (7) – 600e^{– 7x} . (-7)]

= 49[500e^7x + 600e^{– 7x}]

= 49y

Hence Proved!!

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

Solution:

e^y (x + 1) = 1

e^-y = (x+1)

First derivative,

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

-e^-y $\frac{dy}{dx}$ = 1

$\frac{dy}{dx} = \frac{-1}{e^{-y}}$

= $\frac{-1}{(x+1)}$

Second derivative,

$\frac{d^2y}{dx^2} = \frac{d}{dx}(\frac{dy}{dx})$

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

Using division rule,

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

= $\frac{(x+1) . (0) + (1)}{(x+1) ^2}$

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

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

Hence we can conclude that,

$\mathbf{\frac{d^2y}{dx^2} = (\frac{dy}{dx})^2}$

Question 17. If y = (tan^–1 x)², show that (x²+ 1)² y₂+ 2x (x²+ 1) y₁= 2

Solution:

Here, y = (tan–1 x)²

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

= 2 . tan–1 x $\frac{1}{x^2 + 1}$

(x² + 1) $\frac{dy}{dx}$ = 2 tan–1 x

Derivation further,

(x² + 1) $\frac{d^2y}{dx^2}$ + $\frac{dy}{dx} \frac{d}{dx}$ (x2 + 1) = $\frac{d}{dx}(2 tan^{-1} x)$

(x² + 1) $\frac{d^2y}{dx^2}$ + $\frac{dy}{dx}$ (2x) = 2 $\frac{1}{x^2+1}$

Multiplying (x² + 1),

(x² + 1)² $\frac{d^2y}{dx^2}$ + $\frac{dy}{dx}$ (2x)(x² + 1) = 2

Hence Proved,

(x²+ 1)² y₂+ 2x (x²+ 1) y₁= 2