These NCERT Solutions for Class 12 Maths Chapter 5 Continuity and Differentiability Ex 5.5 Questions and Answers are prepared by our highly skilled subject experts.

## NCERT Solutions for Class 12 Maths Chapter 5 Continuity and Differentiability Exercise 5.5

Question 1.

cos x. cos 2x. cos 3x

Solution:

Let y = cos x. cos 2x . cos 3x,

Taking log on both sides,

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

Question 2.

\(\sqrt{\frac{(x-1)(x-2)}{(x-3)(x-4)(x-5)}}\)

Solution:

Question 3.

(log x)^{cosx}

Solution:

let y = (log x)^{cosx}

Taking log on both sides,

log y = log (log x)^{cosx}

log y = cos x log (log x),

Differentiating w.r.t. x,

Question 4.

x – 2^{sinx}

Solution:

let y = x – 2^{sinx},

∴ y = u – v

Differentiating w.r.t. x,

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,

logy = 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

On simplification,

\(\frac { dy }{ dx }\) = (x + 3)(x + 4)²(x + 5)³(9x² + 70x + 133)

Question 6.

\({ \left( x+\frac { 1 }{ x } \right) }^{ x }+{ x }^{ \left( 1+\frac { 1 }{ x } \right) }\)

Solution:

Let u = \(\left(x+\frac{1}{x}\right)^{x}\) = \(\left(\frac{x^{2}+1}{x}\right)^{x}\)

v = \(x^{\left(1+\frac{1}{x}\right)}\)

Let y = u + v

Differentiating w.r.t. x,

∴ \(\frac { dy }{ dx }\) = \(\frac { du }{ dx }\) + \(\frac { dv }{ dx }\) … (1)

u = \(\left(\frac{x^{2}+1}{x}\right)^{x}\)

Taking logaritham on both sides,

logu = x log \(\left(\frac{x^{2}+1}{x}\right)\)

∴ log u = x(log(x²+ 1) – logx)

Differentiating both sides w.r.t.x,

v = \(x^{\left(1+\frac{1}{x}\right)}\)

Taking logaritham on both sides,

logv = (1 + \(\frac { 1 }{ x }\)) log x

Differentiating both sides w.r.t. x,

Question 7.

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

Solution:

Let u = (log x)^{x}, v = x^{logx} and y = u + v

Differentiating w.r.t. x,

∴ \(\frac { dy }{ dx }\) = \(\frac { du }{ dx }\) + \(\frac { dv }{ dx }\) … (1)

u = (log x)^{x}

Taking logaritham on both sides,

logu = x log(log x)

Differentiating both sides w.r.t. x,

v = x^{log x}

Taking logarithm on both sides,

logv = logx. logx = (logx)²

Differentiating both sides w.r.t. x,

Question 8.

(sin x)^{x}+sin^{-1} \(\sqrt{x}\)

Solution:

Let y = (sin x)^{x }+ sin^{-1 }\(\sqrt{x}\)

Let y = u + v

Differentiating w.r.t. x,

∴ \(\frac { dy }{ dx }\) = \(\frac { du }{ dx }\) + \(\frac { dv }{ dx }\) … (1)

u = (sin x)^{x}

\(\frac { du }{ dx }\) = (sin x)^{x}[x cotx + log sinx] … (2)

Let y = (sinx)^{x}

Taking logarithm on bath sides,

logy = x.log sin x

Differentiating both sides w.r.t. x,

Question 9.

x^{sinx} + (sin x)^{cosx}

Solution:

Let y = x^{sinx} + (sin x)^{cosx}

Differentiating both sides w.r.t. x,

(i) Let y = (sin x)^{x}

Taking logarithm on both sides,

logy = sinx. logx

Differentiating both sides w.r.t. x,

(ii) Let y = (sinx)^{cosx }

Taking logarithm on both sides,

logy = cosx. log sinx

Differentiating both sides w.r.t. x,

Question 10.

\({ x }^{ x\quad cosx }+\frac { { x }^{ 2 }+1 }{ { x }^{ 2 }-1 } \)

Solution:

Let u = x^{x cosx} and v = \(\frac{x^{2}+1}{x^{2}-1}\)

Let y = u + v

Differentiating w.r.t. x,

∴ \(\frac { dy }{ dx }\) = \(\frac { du }{ dx }\) + \(\frac { dv }{ dx }\) … (1)

u = x^{x cosx}

Taking logarithm on both sides,

logu = x cosx . logx

Differentiating both sides w.r.t. x,

Differentiating both sides w.r.t. x,

Question 11.

\((x \cos x)^{x}+(x \sin x)^{\frac{1}{x}}\)

Solution:

Let u = (x cosx)^{x} and v = \((x \sin x)^{\frac{1}{x}}\)

u = (x cosx)^{x}

Taking logarithm on both sides,

log u = x log (x cosx)

log u = x[log x + log cosx]

Differentiating w.r.t. x,

Question 12.

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

Solution:

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

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

∴ u + v = 1,

\(\frac { du }{ dx } +\frac { dv }{ dx }=0\)

Now u = x^{y}

Taking logarithm on both sides,

log u = y log x

Differentiating w.r.t. x

Question 13.

y^{x} = x^{y}

Solution:

x^{y} = y^{x}

Taking logarithm on both sides,

y log x = x log y

Differentiating w.r.t. x

Question 14.

(cos x)^{y} = (cos y)^{x}

Solution:

(cos x)^{y} = (cos y)^{x}

Taking logarithm on both sides,

y log cosx = x log cosy

Differentiating both sides w.r.t. x,

Question 15.

xy = e^{(x-y)}

Solution:

log(xy) = log e^{(x-y)}

⇒ log(xy) = x – y

⇒ logx + logy = x – y

\( ⇒\frac { 1 }{ x } +\frac { 1 }{ y } \frac { dy }{ dx } =1-\frac { dy }{ dx } ⇒\frac { dy }{ dx } =\frac { y(x-1) }{ x(y+1) } \)

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

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

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

Differentiating w.r.t. x,

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

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

f = 5x^{4} – 20x^{3} + 45x^{2} – 52x + 11.

(ii) y = x^{5} – 5x^{4} + 15x³ – 26x² + 11x + 72

(by expanding the product)

Differentiating w.r.t. x, dy

\(\frac { dy }{ dx }\) = 5x^{4} – 20x³ + 45x² – 52x + 11

(iii) y = (x² – 5x + 8)(x³ + 7x + 9)

Taking logarithm on both sides,

log y = log(x² – 5x 4- 8)(x³ + 7x + 9)

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

Differentiating, w.r.t x,

= (2x – 5) (x³ + 7x + 9) + (3x² + 7) (x² – 5x + 8)

= 5x^{4} – 20x³ + 45x² – 52x + 11

Yes, the answers are the same.

Question 18.

If u, v and w are functions of w then show that

\(\frac { d }{ dx } (u.v.w)=\frac { du }{ dx } v.w+u.\frac { dv }{ dx } .w+u.v\frac { dw }{ dx } \)

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

Solution:

Let y = u.v.w

(a) Product rule

(b) Logarithmic differentiation

y = uvw

Taking logarithm on both sides, we get

log y = log uvw

i.e., log y = log u + log v + log w

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