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
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.
ex sin 5x
Solution:
Let y = ex sin 5x
∴ (frac { dy }{ dx }) = ex. (frac { d }{ dx })(sin 5x) + sin 5x(frac { d }{ dx })ex
= ex.cos 5x . 5 + sin 5x . ex
= ex(5 cos 5x + sin 5x)
So, (frac{d^{2} y}{d x^{2}}) = ex . (frac { d }{ dx })(5 cos 5x + sin 5x) + (5 cos 5x + sin 5x) . (frac { d }{ dx })(ex)
= ex(- 25 sin 5x + 5 cos 5x) + (5 cos 5x + sin 5x) ex
= ex(- 24 sin 5x + 10 cos 5x)
= 2ex (5 cos 5x – 12 sin 5x).
Question 7.
e6x cos 3x
Solution:
Let y = e6x cos 3x
∴ (frac { dy }{ dx }) = e6x. (frac { d }{ dx })(cos 5x) + cos 3x (frac { d }{ dx })e6x
= e6x(- 3 sin 3x (6 e6x)
= e6x(- 3 sin 3x + 6 cos 3x)
So, (frac{d^{2} y}{d x^{2}}) = e6x (frac { d }{ dx })(- 3 sin 3x + 6 cos 3x) + (- 3 sin 3x + 6 cos 3x) (frac { d }{ dx })e6x
= e6x(- 9 cos 3x – 18 sin 3x) + (- 3 sin 3x + 6 cos 3x) e6x . 6
= 9 e6x(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 = Aemx + Benx, show that (frac{d^{2} y}{d x^{2}}) – (m + n)(frac { dy }{ dx }) + mny = 0.
Solution:
We have : y = Aemx + Benx.
∴ (frac { dy }{ dx }) = A . memx + B . nenx
So, (frac{d^{2} y}{d x^{2}}) = A . m²emx + B . n² enx.
∴ (frac{d^{2} y}{d x^{2}}) – (m+n) (frac { dy }{ dx }) + mny
= A m² emx + B n² enx – (m + n) x (A memx + B nenx) + mn (A emx + B enx)
= A m² emx + B n² enx – A m² emx – B mnenx – A mnemx – B n² enx + A mnemx + B mnenx
= 0.
Question 15.
If y = 500 e7x + 600 e-7x, show that (frac{d^{2} y}{d x^{2}}) = 49y.
Solution:
We have:
Question 16.
If ey (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.