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

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.