NCERT Solutions for Class 12 Maths Chapter 5 Continuity and Differentiability Ex 5.5 are part of NCERT Solutions for Class 12 Maths. Here we have given NCERT Solutions for Class 12 Maths Chapter 5 Continuity and Differentiability Ex 5.5.

- Continuity and Differentiability Class 12 Ex 5.1
- Continuity and Differentiability Class 12 Ex 5.2
- Continuity and Differentiability Class 12 Ex 5.3
- Continuity and Differentiability Class 12 Ex 5.4
- Continuity and Differentiability Class 12 Ex 5.6
- Continuity and Differentiability Class 12 Ex 5.7
- Continuity and Differentiability Class 12 Ex 5.8

Board |
CBSE |

Textbook |
NCERT |

Class |
Class 12 |

Subject |
Maths |

Chapter |
Chapter 5 |

Chapter Name |
Continuity and Differentiability |

Exercise |
Ex 5.5 |

Number of Questions Solved |
18 |

Category |
NCERT Solutions |

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

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

**Ex 5.5 Class 12 Maths 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

**Ex 5.5 Class 12 Maths Question 2.**

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

**Solution:**

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

taking log on both sides

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

**Ex 5.5 Class 12 Maths 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,

**Ex 5.5 Class 12 Maths Question 4.**

x – 2^{sinx}

**Solution:**

let y = x – 2^{sinx},

y = u – v

**Ex 5.5 Class 12 Maths 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

**Ex 5.5 Class 12 Maths Question 6.**

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

**Solution:**

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

let \(u={ \left( x+\frac { 1 }{ x } \right) }^{ x }and\quad v={ x }^{ \left( 1+\frac { 1 }{ x } \right) }\)

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

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

**Solution:**

let y = (log x)^{x} + x^{logx} = u+v

where u = (log x)^{x}

∴ log u = x log(log x)

**Ex 5.5 Class 12 Maths Question 8.**

(sin x)^{x}+sin^{-1} √x

**Solution:**

Let y = (sin x)^{x }+ sin^{-1 }√x

let u = (sin x)x, v = sin^{-1} √x

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

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

**Solution:**

let y = x^{sinx} + (sin x)^{cosx} = u+v

where u = x^{sinx}

log u = sin x log x

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

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

**Solution:**

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

y = u + v

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

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

**Solution:**

\(y={ (x\quad cosx) }^{ x }+{ (x\quad sinx) }^{ \frac { 1 }{ x } }\)

Let u = (x cosx)^{x}

logu = x log(x cosx)

**Find \(\\ \frac { dy }{ dx } \) of the functions given in Questions 12 to 15.**

**Ex 5.5 Class 12 Maths 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

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

y^{x }= x^{y}

**Solution:**

y = x

x logy = y logx

**Ex 5.5 Class 12 Maths Question 14.**

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

**Solution:**

We have

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

=> y log (cosx) = x log (cosy)

**Ex 5.5 Class 12 Maths Question 15.**

xy = e^{(x-y)}

**Solution:**

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

=> log(xy) = x – y

=> logx + logy = x – y

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

**Ex 5.5 Class 12 Maths 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})

**Ex 5.5 Class 12 Maths 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) By expanding the product to obtain a single polynomial, we get

**Ex 5.5 Class 12 Maths 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

=> y = u. (vw)

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