NCERT Solutions for Maths Class 12 Exercise 5.5

# NCERT Solutions for Maths Class 12 Exercise 5.5

Hello Students! In this post, you will find the complete NCERT Solutions for Maths Class 12 Exercise 5.5.

## NCERT Solutions for Maths Class 12 Exercise 5.5

Differentiate the functions given in Questions 1 to 11 w.r.t. x.

Maths Class 12 Ex 5.5 Question 1.
cos x. cos 2x. cos 3x

Solution:
Let y = cos x. cos 2x . cos 3x
Taking log on both sides, we get
log y = log (cos x. cos 2x. cos 3x)
log y = log cos x + log cos 2x + log cos 3x
Differentiating w.r.t. x, we get

Maths Class 12 Ex 5.5 Question 2.

Solution:

Taking log on both sides, we get

Differentiating both sides w.r.t. x, we get

Maths Class 12 Ex 5.5 Question 3.
(log x)cos x

Solution:
Let y = (log x)cos x
Taking log on both sides, we get
log y = log (log x)cos x
log y = cos x log (log x)
Differentiating both sides w.r.t. x, we get

Maths Class 12 Ex 5.5 Question 4.
xx – 2sin x

Solution:
Let y = xx – 2sin x
y = u – v, where u = xx and v = 2sin x

Maths Class 12 Ex 5.5 Question 5.
(x + 3)2 . (x + 4)3 . (x + 5)4

Solution:
Let y = (x + 3)2 . (x + 4)3 . (x + 5)4
Taking log on both side, we get
log y = log [(x + 3)2 . (x + 4)3 . (x + 5)4]
= log (x + 3)2 + log (x + 4)3 + log (x + 5)4
log y = 2 log (x + 3) + 3 log (x + 4) + 4 log (x + 5)
Differentiating w.r.t. x, we get

Maths Class 12 Ex 5.5 Question 6.

Solution:

Differentiating both sides w.r.t. x, we get

Taking log on both sides, we get

Differentiating both sides w.r.t. x, we get

Maths Class 12 Ex 5.5 Question 7.
(log x)x + xlog x

Solution:
Let y = (log x)x + xlog x = u + v
Where u = (log x)x
Taking log on both sides, we get

log u = x log(log x)

Differentiating both sides w.r.t. x, we get

Maths Class 12 Ex 5.5 Question 8.

(sin x)x + sin-1 √x

Solution:
Let y = (sin x)+ sin-1 √x
Let u = (sin x)x and v = sin-1 √x

Maths Class 12 Ex 5.5 Question 9.
xsin x + (sin x)cos x

Solution:
Let y = xsin x + (sin x)cos x = u + v
Where u = xsin x
log u = sin x log x

Maths Class 12 Ex 5.5 Question 10.

Solution:

Taking log on both sides, we get

log u = log (xxcos x)

log u = x cos x log x

Differentiating both sides w.r.t. x, we get

Taking log on both sides, we get

log v = log (x2 + 1) – log (x2 − 1)

Differentiating both sides w.r.t. x, we get

Maths Class 12 Ex 5.5 Question 11.

Solution:

Putting the values from equations (ii) and (iii) in equation (i), we get

Find dy/dx of the functions given in Questions 12 to 15.

Maths Class 12 Ex 5.5 Question 12.
xy + yx = 1

Solution:
We have, xy + yx = 1
Let u = xy and v = yx
u + v = 1,

Now, u = xy

Taking log on both sides, we get

log u = log xy

log u = y log x

Differentiating both sides w.r.t. x, we get

Now, v = yx

Taking log on both sides, we get

log v = log yx

log v = x log y

Differentiating both sides w.r.t. x, we get

Maths Class 12 Ex 5.5 Question 13.
y= xy

Solution:
We have, y= xy
Taking log on both sides, we get
x log y = y log x

Differentiating both sides w.r.t. x, we get

Maths Class 12 Ex 5.5 Question 14.

(cos x)y = (cos y)x

Solution:
We have, (cos x)y = (cos y)x
Taking log on both sides, we get

y log (cos x) = x log (cos y)

Differentiating both sides w.r.t. x, we get

Maths Class 12 Ex 5.5 Question 15.

xy = e(x - y)

Solution:
We have, xy = e(x - y)
Taking log on both sides, we get

log (xy) = log e(x - y)
log (xy) = (x – y) log e
log x + log y = x – y                        (Since log e = 1)

Differentiating both sides w.r.t. x, we get

Maths Class 12 Ex 5.5 Question 16.

Find the derivative of the function given by f(x) = (1 + x) (1 + x2) (1 + x4) (1 + x8) and hence find f'(1).

Solution:
Let f(x) = y = (1 + x) (1 + x2) (1 + x4) (1 + x8)
Taking log both sides, we get
log y = log [(1 + x) (1 + x2) (1 + x4) (1 + x8)]
log y = log(1 + x) + log (1 + x2) + log(1 + x4) + log(1 + x8)

Maths Class 12 Ex 5.5 Question 17.

Differentiate (x2 – 5x + 8) (x3 + 7x + 9) in three ways mentioned below:
(i) by using product rule
(ii) by expanding the product to obtain a single polynomial.
(iii) by logarithmic differentiation.
Do they all give the same answer?

Solution:
(i) By using product rule,

Let y = (x2 – 5x + 8) (x3 + 7x + 9)

Let u = (x2 – 5x + 8) and v = (x3 + 7x + 9)

(ii) By expanding the product to obtain a single polynomial, we get

y = (x− 58(x79)

= x2(x79− 5(x798(x79)

= x7x9x− 5x− 35x− 458x5672

y = x− 5x15x− 26x1172
Differentiating both sides w.r.t. x, we get

(iii) By logarithmic differentiation

y = (x2 − 5x + 8) (x3 + 7x + 9)

Taking log on both the sides, we get

log y = log (x2 − 5x + 8) + log (x3 + 7x + 9)

Differentiating both sides with respect to x, we get

From the above three observations, it can be seen that all the results of dy/dx

Maths Class 12 Ex 5.5 Question 18.
If u, v and w are functions of x, then show that

in two ways - first by repeated application of product rule, second by logarithmic differentiation.

Solution:
Let y = u.v.w
y = u. (v.w)
By applying product rule, we get

Taking logarithm on both the sides of the equation y = u.v.w, we get

log y = log u + log v + log w

Differentiating both sides w.r.t. x, we get