NCERT Solutions Class 12 maths Chapter-5 (Continuity And Differentiability)Exercise 5.7

NCERT Solutions Class 12 Maths from class 12th Students will get the answers of Chapter-5 (Continuity And Differentiability)Exercise 5.7 This chapter will help you to learn the basics and you should expect at least one question in your exam from this chapter.
We have given the answers of all the questions of NCERT Board Mathematics Textbook in very easy language, which will be very easy for the students to understand and remember so that you can pass with good marks in your examination.

Exercise 5.7

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,

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

= 20x20-1

= 20x19

Second derivative,

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

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

= 20(19x19-1)

= 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 = ex sin 5x

First derivative,

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

Using product rule

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

= ex (5 cos(5x))+ sin 5x (ex)

= ex (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,

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

= ex (5 (5(- sin 5x))) + 5(cos 5x) + (5 cos(5x)+ sin 5x) (ex)

= ex (- 25 sin 5x + 5cos 5x) + (5 cos(5x)+ sin 5x) (ex)

= ex (- 25 sin 5x + 5cos 5x + 5 cos(5x)+ sin 5x)

= ex (10 cos 5x – 24 sin 5x)

Question 7. e^6x cos 3x

Solution:

Here, y = e6x cos 3x

First derivative,

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

Using product rule

= e6x $\frac{d(cos (3x))}{dx}$ + cos 3x $\frac{d(e^{6x})}{dx}$

= e6x (- 3 sin(3x))+ cos 3x (6e6x)

= e6x (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,

= e6x ( $\frac{d(6 cos(3x) - 3 sin (3x))}{dx}$ ) + (6 cos(3x) – 3 sin (3x)) $\frac{d(e^{6x})}{dx}$

= e6x (6 (3 (- sin(3x)) – 3 (3 cos 3x)) + (6 cos(3x) – 3 sin (3x)) (6e6x)

= e6x (- 18sin(3x) – 9 cos 3x) + (36 cos(3x) – 18 sin (3x)) (e6x)

= e6x (27 cos(3x) – 36 sin (3x))

= 9e6x (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 + 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 = Aemx + Benx

First derivative,

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

= mAemx + nBenx

Second derivative,

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

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

= m2Aemx + n2Benx

According to the given conditions,

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

LHS = m2Aemx + n2Benx – (m+n)(mAemx + nBenx) + mny

= m2Aemx + n2Benx – (m2Aemx + mnAemx + mnBenx + n2Benx) + mny

= -(mnAemx + mnBenx) + 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 = 500e7x+ 600e– 7x

First derivative,

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

= 500e7x . (7)+ 600e– 7x (-7)

= 7(500e7x – 600e– 7x)

Second derivative,

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

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

= 7[500e7x . (7) – 600e– 7x . (-7)]

= 49[500e7x + 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:

ey (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

NCERT Solutions Class 12 maths Chapter-5 (Continuity And Differentiability)Exercise 5.7

NCERT Solutions Class 12 maths Chapter-5 (Continuity And Differentiability)Exercise 5.7

Exercise 5.7

Chapter-5 (Continuity And Differentiability)

Exercise 5.1
Exercise 5.2
Exercise 5.3
Exercise 5.4
Exercise 5.5
Exercise 5.6
Exercise 5.8

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NCERT Solutions Class 12 maths Chapter-5 (Continuity And Differentiability)Exercise 5.7

NCERT Solutions Class 12 maths Chapter-5 (Continuity And Differentiability)Exercise 5.7

Exercise 5.7

Chapter-5 (Continuity And Differentiability)

Exercise 5.1Exercise 5.2Exercise 5.3Exercise 5.4Exercise 5.5Exercise 5.6Exercise 5.8

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Exercise 5.1
Exercise 5.2
Exercise 5.3
Exercise 5.4
Exercise 5.5
Exercise 5.6
Exercise 5.8