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

You can download the **PDF of NCERT Books Maths Chapter 5** for your easy reference while studying **NCERT Solutions for Maths Class 12 Exercise 5.5**.

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**NCERT Solutions for Maths Class 12 Exercise 5.1**

**NCERT Solutions for Maths Class 12 Exercise 5.2**

**NCERT Solutions for Maths Class 12 Exercise 5.3**

**NCERT Solutions for Maths Class 12 Exercise 5.4**

**NCERT Solutions for Maths Class 12 Exercise 5.6**

**NCERT Solutions for Maths Class 12 Exercise 5.7**

**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

**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.**

x^{x} – 2^{sin
x}

**Solution:**

Let y = x^{x}
– 2^{sin x}

y = u – v, where u = x^{x} and v = 2^{sin 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.**

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

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

**Maths Class
12 Ex 5.5 Question 7.**

(log x)^{x} +
x^{log x}

**Solution:**

Let y = (log x)^{x} +
x^{log 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

(sin x)^{x }+
sin^{-1} √x

**Solution:**

Let y = (sin x)^{x }+
sin^{-1 }√x

Let u = (sin x)^{x} and v = sin^{-1} √x

**Maths Class
12 Ex 5.5 Question 9.**

x^{sin x} +
(sin x)^{cos x}

**Solution:**

Let y = x^{sin
x} + (sin x)^{cos x} = u + v

Where u = x^{sin x}

log u = sin x log x

**Maths Class
12 Ex 5.5 Question 10.**

Taking log on both sides, we get

⇒ log u = log (x^{xcos x})

⇒ log u = x cos x log x

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

⇒ log v = log (x^{2} + 1)
– log (x^{2} − 1)

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

**Maths Class
12 Ex 5.5 Question 11.**

**Solution:**

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

**Maths Class
12 Ex 5.5 Question 12.**

x^{y} +
y^{x} = 1

**Solution:**

We have, x^{y} +
y^{x} = 1

Let u = x^{y} and v = y^{x}

∴ u + v = 1,

Now, u = x^{y}

Taking log on
both sides, we get

log u = log x^{y}

log u = y log x^{}

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

^{x}

Taking log on both sides, we get

log v = log y^{x}

log v = x log y

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

**Maths Class
12 Ex 5.5 Question 13.**

y^{x }=
x^{y}

**Solution:**

We have, y^{x }=
x^{y}

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 + x^{2}) (1 + x^{4})
(1 + x^{8}) and hence find f'(1).

**Solution:**

Let f(x) = y = (1
+ x) (1 + x^{2}) (1 + x^{4}) (1 + x^{8})

Taking log both sides, we get

log y = log [(1 + x) (1 + x^{2}) (1 + x^{4}) (1 + x^{8})]

log y = log(1 + x) + log (1 + x^{2}) + log(1 + x^{4}) + log(1 +
x^{8})

**Maths Class 12 Ex 5.5 Question 17.**

Differentiate (x^{2} – 5x + 8) (x^{3} + 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 = (x^{2} –
5x + 8) (x^{3} + 7x + 9)

Let u = (x^{2} –
5x + 8) and v = (x^{3} + 7x + 9)

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

y = (x^{}2 − 5x + 8) (x^{}3 + 7x + 9)

= x^{}2(x^{}3 + 7x + 9) − 5x (x^{}3 + 7x + 9) + 8(x^{}3 + 7x + 9)

= x^{}5 + 7x^{}3 + 9x^{}2 − 5x^{}4 − 35x^{}2 − 45x + 8x^{}3 + 56x + 72

y = x^{}5 − 5x^{}4 + 15x^{}3 − 26x^{}2 + 11x + 72

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

**(iii)** By
logarithmic differentiation

y = (x^{2} − 5x + 8) (x^{3}
+ 7x + 9)

Taking log on both the
sides, we get

log y = log (x^{2} − 5x
+ 8) + log (x^{3} + 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

log y = log u + log v + log w

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

**Related Links:**

**NCERT Solutions for Maths Class 12 Exercise 5.1**

**NCERT Solutions for Maths Class 12 Exercise 5.2**

**NCERT Solutions for Maths Class 12 Exercise 5.3**

**NCERT Solutions for Maths Class 12 Exercise 5.4**