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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}



= 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. x3 log 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. ex sin 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. e6x 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

Solution:

Here, y = tan–1

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 = Aemx + Benx, 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 = 500e7x+ 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 ey (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

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