# NCERT Solutions For Class 12 Maths Chapter 5 Continuity and Differentiability Ex 5.5

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Textbook | NCERT |

Board | CBSE |

Category | NCERT Solutions |

Class | Class 12 |

Subject | Maths |

Chapter | Chapter 5 |

Exercise | Class 12 Maths Chapter 5 Continuity and Differentiability Exercise 5.5 |

Number of Questions Solved | 18 |

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

**NCERT TEXTBOOK EXERCISES**

**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 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)

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

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### Other Chapter of Class 12 Maths Chapter 5 Continuity and Differentiability

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