GSEB Solutions Class 12 Maths Chapter 5 Continuity and Differentiability Ex 5.7

Gujarat Board GSEB Textbook Solutions Class 12 Maths Chapter 5 Continuity and Differentiability Ex 5.7 Textbook Questions and Answers.

Gujarat Board Textbook Solutions Class 12 Maths Chapter 5 Continuity and Differentiability Ex 5.7

Question 1.
x² + 3x + 2
Solution:
∴ $\frac { dy }{ dx }$ = 2x + 3
So, $\frac{d^{2} y}{d x^{2}}$ = $\frac { d }{ dx }$(2x + 3) = 2.

Question 2.
x²⁰
Solution:
∴ $\frac { dy }{ dx }$ = x²⁰
So, $\frac{d^{2} y}{d x^{2}}$ = 20 $\frac { d }{ dx }$(x¹⁹) = 20 x 19 x¹⁸ = 380 x¹⁸.

Question 3.
x cos x
Solution:
Let y = x cos x.
∴ $\frac { dy }{ dx }$ = x $\frac { d }{ dx }$ (cos x) + cos x $\frac { d }{ dx }$(x).
= x(- sin x) + cos x . 1
= – x sin x + cos x.
So, $\frac{d^{2} y}{d x^{2}}$ = – $\frac { d }{ dx }$(x sin x) + $\frac { d }{ dx }$ (cos x)
= – (x cos x + sin x . 1) – sin x
= – x cos x – 2 sin x.

Question 4.
log x
Solution:
Let y = log x
∴ $\frac { dy }{ dx }$ = $\frac { 1 }{ x }$ = x^-1
So, $\frac{d^{2} y}{d x^{2}}$ = $\frac { d }{ dx }$(x^-1) = – 1. x^-2 = – $\frac{1}{x^{2}}$

Question 5.
x³ log x
Solution:
Let y = x³ log x

Question 6.
e^x sin 5x
Solution:
Let y = e^x sin 5x
∴ $\frac { dy }{ dx }$ = e^x. $\frac { d }{ dx }$(sin 5x) + sin 5x$\frac { d }{ dx }$e^x
= e^x.cos 5x . 5 + sin 5x . e^x
= e^x(5 cos 5x + sin 5x)
So, $\frac{d^{2} y}{d x^{2}}$ = e^x . $\frac { d }{ dx }$(5 cos 5x + sin 5x) + (5 cos 5x + sin 5x) . $\frac { d }{ dx }$(e^x)
= e^x(- 25 sin 5x + 5 cos 5x) + (5 cos 5x + sin 5x) e^x
= e^x(- 24 sin 5x + 10 cos 5x)
= 2e^x (5 cos 5x – 12 sin 5x).

Question 7.
e^6x cos 3x
Solution:
Let y = e^6x cos 3x
∴ $\frac { dy }{ dx }$ = e^6x. $\frac { d }{ dx }$(cos 5x) + cos 3x $\frac { d }{ dx }$e^6x
= e^6x(- 3 sin 3x (6 e^6x)
= e^6x(- 3 sin 3x + 6 cos 3x)
So, $\frac{d^{2} y}{d x^{2}}$ = e^6x $\frac { d }{ dx }$(- 3 sin 3x + 6 cos 3x) + (- 3 sin 3x + 6 cos 3x) $\frac { d }{ dx }$e^6x
= e^6x(- 9 cos 3x – 18 sin 3x) + (- 3 sin 3x + 6 cos 3x) e^6x . 6
= 9 e^6x(3 cos 3x – 4 sin 3x).

Question 8.
tan^-1 x
Solution:
∴ $\frac { dy }{ dx }$ = $\frac{1}{1+x^{2}}$ = (1 + x²)^-1.
So, $\frac{d^{2} y}{d x^{2}}$ = $\frac { dy }{ dx }$ (1 + x²)^-1.
= (-1) . (1 + x²)^-2 . 2x = $\frac{-2 x}{\left(1+x^{2}\right)^{2}}$.

Question 9.
log (log x)
Solution:
Let y = log (log x)

Question 10.
sin (log x)
Solution:
Let y = sin (log x).

Question 11.
If y = 5 cos x – 3 sin x, prove that $\frac{d^{2} y}{d x^{2}}$ + y = 0.
Solution:
We have: y = 5 cos x – 3 sin x.
∴ $\frac { dy }{ dx }$ = – 5 sin x – 3 cos x.
So, $\frac{d^{2} y}{d x^{2}}$ = – 5 cos x + 3 sin x = – (5 cos x – 3 sin x) = – y.
⇒ $\frac{d^{2} y}{d x^{2}}$ + y = 0.

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

Question 13.
If y = 3 cos (log x) + 4 sin (log x), show that x²y₂ + xy₁ + y = 0.
Solution:

Question 14.
If y = Ae^mx + Be^nx, show that $\frac{d^{2} y}{d x^{2}}$ – (m + n)$\frac { dy }{ dx }$ + mny = 0.
Solution:
We have : y = Ae^mx + Be^nx.
∴ $\frac { dy }{ dx }$ = A . me^mx + B . ne^nx
So, $\frac{d^{2} y}{d x^{2}}$ = A . m²e^mx + B . n² e^nx.
∴ $\frac{d^{2} y}{d x^{2}}$ – (m+n) $\frac { dy }{ dx }$ + mny
= A m² e^mx + B n² e^nx – (m + n) x (A me^mx + B ne^nx) + mn (A e^mx + B e^nx)
= A m² e^mx + B n² e^nx – A m² e^mx – B mne^nx – A mne^mx – B n² e^nx + A mne^mx + B mne^nx
= 0.

Question 15.
If y = 500 e^7x + 600 e^-7x, show that $\frac{d^{2} y}{d x^{2}}$ = 49y.
Solution:
We have:

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

Question 17.
If y = (tan^-1 x)², show that (x² + 1)² y₂ + 2x (x² + 1) y₁ = 2.
Solution:
We have : y = (tan^-1 x)².
∴ y₁ = 2 tan^-1x $\frac { d }{ dx }$(tan^-1x),
⇒ y₁ = $\frac{2 \tan ^{-1} x}{1+x^{2}}$
⇒ (1 + x²)y₁ = 2 tan^-1x.
Differentiating again w.r.t. x, we have :
(1 + x²)y₂ + y₁(2x) = $\frac{2}{1+x^{2}}$
⇒ (1 + x²)y₂ + 2x(1 + x²)y₁ = 2.