AP Inter 2nd Year Maths Exercise 5g Solutions

Referring to the AP Inter 2nd Year Maths Study Material Chapter 5 Continuity and Differentiability Exercise 5g Solutions makes it easier to understand complex problems.

AP Inter 2nd Year Maths Continuity and Differentiability Solutions Exercise 5g

I.

Question 1.
Find the second order derivative of x² + 3x + 2
Solution:
Let y = x² + 3x + 2
∴ \(\frac{\mathrm{dy}}{\mathrm{dx}}\) = 2x + 3 . 1 + 0 = 2x + 3.
Again differentiating w.r.t. x, we have
\(\frac{\mathrm{d}^2 \mathrm{y}}{\mathrm{dx}^2}=\frac{d}{d x}\left(\frac{d y}{d x}\right)\) = 2(1) + 0 = 2

Question 2.
Find the second order derivative of x²⁰.
Solution:
Let y = x²⁰
∴ \(\frac{\mathrm{dy}}{\mathrm{dx}}\) = 20x¹⁹
Again differentiating w.r.t. x, we have
\(\frac{\mathrm{d}^2 \mathrm{y}}{\mathrm{dx}^2}\) = 20 × 19x¹⁸ = 380x¹⁸

Question 3.
Find the second order derivative of x . cos x
Solution:
Let y = x cos x
∴ \(\frac{\mathrm{dy}}{\mathrm{dx}}\) = x\(\frac{\mathrm{d}}{\mathrm{dx}}\)cos x + c0s x \(\frac{\mathrm{d}}{\mathrm{dx}}\)x [By Product Rule]
= -x sin x + cos x .
Again differentiating w.r.t. x, we have
\(\frac{\mathrm{d}^2 \mathrm{y}}{\mathrm{dx}^2}\) = –\(\frac{\mathrm{d}}{\mathrm{dx}}\)(x sin x) + \(\frac{\mathrm{d}}{\mathrm{dx}}\)cos x
= -[x\(\frac{\mathrm{d}}{\mathrm{dx}}\)sin x + sin x\(\frac{\mathrm{d}}{\mathrm{dx}}\)(x)] – sin x
= -(x cos x + sin x) – sin x = -x cos x – sin x – sin x
= -x cos x – 2 sin x = -(x cos x + 2 sin x).

Question 4.
Find the second order derivative of log x
Solution:
Let y = log x
∴ \(\frac{\mathrm{dy}}{\mathrm{dx}}=\frac{1}{x}\)
Again differentiating w.r.t. x, we have
\(\frac{\mathrm{d}^2 \mathrm{y}}{\mathrm{dx}^2}=\frac{d}{d x}\left(\frac{1}{x}\right)\) = \(\frac{\mathrm{d}}{\mathrm{dx}}\)x^-1
= (-1)x^-2 = \(\frac{-1}{x^2}\)

Question 5.
Find the second order derivative of tan^-1 x
Solution:
Let y = tan^-1 x
∴ \(\frac{\mathrm{dy}}{\mathrm{dx}}\) = \(\frac{1}{1+x^2}\)
Again differentiating w.r.t. x, we have
∴ \(\frac{\mathrm{d}^2 \mathrm{y}}{\mathrm{dx}^2}=\frac{\mathrm{d}}{\mathrm{dx}}\left(\frac{1}{1+\mathrm{x}^2}\right)\) = \(\frac{\left(1+x^2\right) \frac{d}{d x}(1)-1 \frac{d}{d x}\left(1+x^2\right)}{\left(1+x^2\right)^2}\)
= \(\frac{\left(1+x^2\right) 0-(2 x)}{\left(1+x^2\right)^2}\) = \(\frac{-2 x}{\left(1+x^2\right)^2}\)

II.

Question 1.
Find the second order derivative of x³ log x
Solution:
Let y = x³ log x
∴ \(\frac{\mathrm{dy}}{\mathrm{dx}}\) = x³\(\frac{\mathrm{d}}{\mathrm{dx}}\) log x + log x \(\frac{\mathrm{d}}{\mathrm{dx}}\) x³ [By Product Rule]
= x³\(\frac{1}{\mathrm{x}}\) + (log x)3x² = x² + 3x² log x
Again differentiating w.r.t. x, we have
∴ \(\frac{\mathrm{d}^2 \mathrm{y}}{\mathrm{dx}^2}\) = \(\frac{\mathrm{d}}{\mathrm{dx}}\) x² + 3\(\frac{\mathrm{d}}{\mathrm{dx}}\) (x² log x) = 2x + 3[x² \(\frac{\mathrm{d}}{\mathrm{dx}}\) log x + log x\(\frac{\mathrm{d}}{\mathrm{dx}}\)x²]
= 2x + 3(x² . \(\frac{1}{\mathrm{x}}\) + (log x)2x) = 2x + 3(x + 2x log x)
= 2x + 3x + 6x log x = 5x + 6x log x
= x(5 + 6 log x)

Question 2.
Find the second order derivative of e^x sin 5x
Solution:
Let y = e^x sin 5x
∴ \(\frac{\mathrm{dy}}{\mathrm{dx}}\) = e^x\(\frac{\mathrm{d}}{\mathrm{dx}}\)sin 5x + sin 5x\(\frac{\mathrm{d}}{\mathrm{dx}}\)e^x [By Product Rule]
= e^x cos5x\(\frac{\mathrm{d}}{\mathrm{dx}}\)5x + sin5xe^x = e^x cos5 x5 + e^x sin5x
= e^x (5 cos 5x + sin 5x)
Again differentiating w.r.t. x using product rule, we have
∴ \(\frac{\mathrm{d}^2 \mathrm{y}}{\mathrm{dx}^2}\) = e^x \(\frac{\mathrm{d}}{\mathrm{dx}}\)(5 cos 5x + sin 5x) + (5 cos 5x + sin 5x)\(\frac{\mathrm{d}}{\mathrm{dx}}\)e^x
= e^x(5(-sin 5x)5 + (cos 5x)5) + (5 cos 5x + sin 5x)e^x
= e^x(-25 sin 5x + 5 cos 5x + 5 cos 5x + sin 5x)
= e^x(10 cos 5x – 24 sin 5x)
= 2e^x(5 cos 5x – 12 sin 5x).

Question 3.
Find the second order derivative of e^6x cos 3x
Solution:
Let y = e^6x cos 3x
∴ \(\frac{\mathrm{dy}}{\mathrm{dx}}\) = e^6x\(\frac{\mathrm{d}}{\mathrm{dx}}\)cos 3x + cos 3x\(\frac{\mathrm{d}}{\mathrm{dx}}\)e^6x [By Product Rule]
= e^6x (-sin 3x)\(\frac{\mathrm{d}}{\mathrm{dx}}\)(3x) + cos 3x . e^6x\(\frac{\mathrm{d}}{\mathrm{dx}}\)6x
= -e^6x sin 3x . 3 + cos 3x e^6x . 6
⇒ \(\frac{\mathrm{dy}}{\mathrm{dx}}\) = 6e^6x cos 3x – 3e^6x sin 3x …………… (1)
∴ \(\frac{\mathrm{d}^2 \mathrm{y}}{\mathrm{dx}^2}\) = \(\frac{\mathrm{d}}{\mathrm{dx}}\) (6e^6x cos 3x – 3e^6x sin 3x) = 6\(\frac{\mathrm{d}}{\mathrm{dx}}\)(e^6x cos 3x) – 3\(\frac{\mathrm{d}}{\mathrm{dx}}\)(e^6x sin 3x)
= 6[6e^6x cos 3x – 3e^6x sin 3x] – 3[sin 3x\(\frac{\mathrm{d}}{\mathrm{dx}}\)(e^6x) + e^6x\(\frac{\mathrm{d}}{\mathrm{dx}}\)(sin 3x)] [using (1)1
= 36e^6x cos 3x – 18e^6x sin 3x – 3[sin 3xe^6x 6 + e^6x cos 3×3]
= 36e^6x cos 3x – 18e^6x sin 3x – 18e^6x sin 3x – 9e^6x cos3x
= 27 e^6x cos 3x – 36e^6x sin 3x = 9e^6x (3 cos 3x – 4 sin 3x)

Question 4.
Find the second order derivative of log (log x)
Solution:
Let y = log (log x)

Question 5.
Find the second order derivative of sin(log x)
Solution:
Let y = sin(log x)

Question 6.
If y = 5 cos x – 3 sin x, prove that \(\frac{\mathrm{d}^2 \mathrm{y}}{\mathrm{dx}^2}\) + y = 0
Solution:
Given that y = 5 cos x – 3 sin x ……………. (i)
⇒ \(\frac{\mathrm{dy}}{\mathrm{dx}}\) = -5 sin x – 3 cos x
Again differentiating w.r.t. x,
\(\frac{\mathrm{d}^2 \mathrm{y}}{\mathrm{dx}^2}\) = -y = -5 cos x + 3 sin x
= -(5 cos x – 3 sin x) = -y (By (i))
⇒ \(\frac{\mathrm{d}^2 \mathrm{y}}{\mathrm{dx}^2}\) = -y
∴ \(\frac{\mathrm{d}^2 \mathrm{y}}{\mathrm{dx}^2}\) + y = 0

Question 7.
If y = cos^-1x. Find \(\frac{\mathrm{d}^2 \mathrm{y}}{\mathrm{dx}^2}\) in terms of y alone.
Solution:
Given that y = cos^-1x ⇒ x = cos y …………. (i)
∴ \(\frac{\mathrm{dy}}{\mathrm{dx}}\) = \(\frac{-1}{\sqrt{1-x^2}}=\frac{-1}{\sqrt{1-\cos ^2 y}}=\frac{-1}{\sqrt{\sin ^2 y}}=\frac{-1}{\sin y}\) = -cosec y [By (i)]
⇒ \(\frac{\mathrm{dy}}{\mathrm{dx}}\) = -cosec y …………. (ii)
Again differentiating both sides w.r.t. x, we have
\(\frac{\mathrm{d}^2 \mathrm{y}}{\mathrm{dx}^2}\) = –\(\frac{\mathrm{d}}{\mathrm{dx}}\) (cosec y) = -[-cosec y cot y \(\frac{\mathrm{dy}}{\mathrm{dx}}\)]
= cosec y cot y(-cosec y) = -cosec²y cot y.

Question 8.
If y 3 cos (log x) + 4 sin (log x). show that x²y₂ + xy₁ + y = 0
Solution:
Given that y = 3cos(log x)+ 4sin(log x)
∴ \(\frac{\mathrm{dy}}{\mathrm{dx}}\) = (y₁)= -3 sin(logx)\(\frac{\mathrm{d}}{\mathrm{dx}}\)logx + 4cos(log x) \(\frac{\mathrm{d}}{\mathrm{dx}}\) logx
⇒ y₁ =- 3 sin(1ogx) \(\frac{1}{x}\) +4cos(logx). \(\frac{1}{x}\)
⇒ xy₁ = -3 sin(log x) +4 cos(log x)
Again differentiating both sides wrt. x,
\(\frac{\mathrm{d}}{\mathrm{dx}}\) (xy₁) = -3 cos(log x)\(\frac{\mathrm{d}}{\mathrm{dx}}\)log x – 4sin(log x)\(\frac{\mathrm{d}}{\mathrm{dx}}\)log x
x\(\frac{\mathrm{d}}{\mathrm{dx}}\)y₁ + y₁\(\frac{\mathrm{d}}{\mathrm{dx}}\)x = – 3 cos(log x)\(\frac{\mathrm{d}}{\mathrm{dx}}\) – 4 sin(log x)\(\frac{1}{\mathrm{dx}}\) [By Product Rule]
⇒ xy₂ + y₁ = –\(\frac{[3 \cos (\log x)+4 \sin (\log x)]}{x}\)
⇒ x(xy₂ + y₁) = -[3 cos(log x)+ 4 sin(log x)]
⇒ x²y₂ + xy₁ = -y ⇒ x²y₂ + xy₁ + y = 0

Question 9.
If y = Ae^mx + Be^nx, show that \(\frac{\mathrm{d}^2 \mathrm{y}}{\mathrm{dx}^2}\) – (m + n) \(\frac{\mathrm{dy}}{\mathrm{dx}}\) + mny = 0
Solution:
Given that y =Ae^mx + Be^nx …………. (i)
∴ \(\frac{\mathrm{dy}}{\mathrm{dx}}\) = Ae^mx\(\frac{\mathrm{d}}{\mathrm{dx}}\)(mx) + Be^nx\(\frac{\mathrm{d}}{\mathrm{dx}}\)(nx) [∵ \(\frac{\mathrm{d}}{\mathrm{dx}}\) e^f(x) = e^f(x) \(\frac{\mathrm{d}}{\mathrm{dx}}\) f(x)]
⇒ \(\frac{\mathrm{dy}}{\mathrm{dx}}\) = Ame^mx + Bne^nx …………. (ii)
∴ \(\frac{\mathrm{d}^2 \mathrm{y}}{\mathrm{dx}^2}\) = Ame^mxm + Bne^nx.n = Am²e^mx + Bn²e^nx ……………. (iii)
Putting values of y, \(\frac{\mathrm{dy}}{\mathrm{dx}}\) and \(\frac{\mathrm{d}^2 \mathrm{y}}{\mathrm{dx}^2}\) from(i), (ii) and (iii) in
L.H.S. = \(\frac{\mathrm{d}^2 \mathrm{y}}{\mathrm{dx}^2}\) – (m + n)\(\frac{\mathrm{dy}}{\mathrm{dx}}\) + mny
= Am²e^mx + Bn²e^nx – (m + n)(Ame^mx +Bne^nx) + mn(Ae^mx + Be^nx)
= Am²e^mx + Bn²e^nx – Am²e^mx – Bmne^nx – Anme^mx – Bn²e^nx + Amne^mx + Bnme^nx = 0
= R.H.S.

Question 10.
If y = 500e^7x + 600e^-7x, show that \(\frac{\mathrm{d}^2 \mathrm{y}}{\mathrm{dx}^2}\) = 49y
Solution:
Given y = 500e^7x + 600e^-7x …………. (i)
⇒ \(\frac{\mathrm{dy}}{\mathrm{dx}}\) = 500e^7x (7) + 600e^-7x(-7) = 500(7)e^7x – 600(7)e^-7x
Now \(\frac{\mathrm{d}^2 \mathrm{y}}{\mathrm{dx}^2}\) = 500(7)e^7x (7) – 600(7)e^-7x (7)
= 500(49)e^7x + 600(49)e^-7x
= 49[500e^7x + 600e^-7x] = 49 y ………………. [By (I)]
∴ \(\frac{\mathrm{d}^2 \mathrm{y}}{\mathrm{dx}^2}\) = 49y.

Question 11.
If e^y (x + 1) = 1, show that \(\frac{d^2 y}{d x^2}=\left(\frac{d y}{d x}\right)^2\)
Solution:
Given that e^y (x + 1) = 1 ⇒ e^y = \(\frac{1}{x+1}\)
Taking logs of both sides, log e^y = log \(\frac{1}{x+1}\)
⇒ y loge = log 1 – log(x + 1)
⇒ y = -log(x + 1) [∵ log e = 1 and log 1 = 0]

Question 12.
If y = (tan^-1x)², show that (x² + 1)²y₂ + 2x (x² + 1)y₁ = 2
Solution:
Given that y = (tan^-1x)²
⇒ y₁ = 2(tan^-1x)\(\frac{\mathrm{d}}{\mathrm{dx}}\)tan^-1x [∵ \(\frac{\mathrm{d}}{\mathrm{dx}}\) (f(x))ⁿ = n(f(x))^n-1\(\frac{\mathrm{d}}{\mathrm{dx}}\)f(x)]
⇒ y₁ = 2(tan^-1x)\(\frac{1}{1+x^2}\)
⇒ y₁ = \(\frac{2 \tan ^{-1} x}{1+x^2}\)
⇒ (1 + x²)y₁ = 2 tan^-1x
Again differentiating both sides w.r.t. x,
(1 + x²)\(\frac{\mathrm{d}}{\mathrm{dx}}\)y₁ + y₁\(\frac{\mathrm{d}}{\mathrm{dx}}\)(1 + x²) = 2 . \(\frac{1}{1+x^2}\)
⇒ (1 + x²)y₂ + y₁ . 2x = \(\frac{2}{1+x^2}\)
⇒ (x² + 1)²y₂ + 2x(1 + x²)y₁ = 2.