Exercise 5.5

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NCERT Solutions class 12 Continuity & Differentiability

Differentiate the functions with respect to  in Exercise 1 to 5.

1.

Ans. Let   ……….(i)

Taking logs on both sides, we have

 =

  [From eq. (i)]

2.

Ans. Let  =   ……….(i)

Taking logs on both sides, we have

  [From eq. (i)]

3.  

Ans. Let  ……….(i)

Taking logs on both sides, we have

   [By Product rule]

  =

4.

Ans. Let

Putting  and

  ……….(i)

Now,

  =  ……….(ii)

Again, 

  ……….(iii)

Putting the values from eq. (ii) and (iii) in eq. (i),

5.  

Ans. Let  …….(i)

Taking logs on both sides, we have

[From eq. (i)

Differentiate the functions with respect to  in Exercise 6 to 11.

6.

Ans. Let

Putting             and

  ……….(i)

Now

=  ……….(ii)

Again

  ……….(iii)

Putting the values from eq. (ii) and (iii) in eq. (i),

7.

Ans. Let  =   where  and

    ……….(i)

Now

  ……….(ii)

Again

   ……….(iii)

Putting the values from eq. (ii) and (iii) in eq. (i),

8.  

Ans. Let  =  where  and

  ……….(i)

Now

   ……….(ii)

Again 

=

=  ……….(iii)

Putting the values from eq. (ii) and (iii) in eq. (i),

9.

Ans. Let

Putting  and , we get 

   ……….(i)

Now

  =

  …..(ii)

Again

  =

  ……….(iii)

Putting values from eq. (ii) and (iii) in eq. (i),

10.

Ans. Let

Putting  and , we have

   ……….(i)

Now   

  =

  ……….(ii)

Again 

  ……….(iii)

Putting the values from eq. (ii) and (iii) in eq. (i),

11.  

Ans. Let

Putting  and , we have

  ……….(i)

Now

  =

  ……….(ii)

Again

  =

  ……….(iii)

Putting the values from eq. (ii) and (iii) in eq. (i)

,

Find  in the following Exercise 12 to 15

12.

Ans. Given:

  where  and

  ……….(i)

Now

  ……….(ii)

Again

 ……….(iii)

Putting values from eq. (ii) and (iii) in eq. (i),

13.

Ans. Given:

14.  

Ans. Given:

15.  

Ans. Given:

16. Find the derivative of the function given by  and hence  

Ans. Given:      ……….(i)

Putting the value of  from eq. (i),

  = 8 x 15 = 120

17. Differentiate  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?

Ans. Let  ……….(i)

(i)

  ……….(ii)

(ii)

  ……….(iii)

(iii)

  [From eq. (i)]

  ……….(iv)

From eq. (ii), (iii) and (iv), we can say that value of  is same obtained by three different methods.

18. If  and  are functions of  then show that  in two ways – first by repeated application of product rule, second by logarithmic differentiation.

Ans. Given:  and  are functions of

To prove:

(i) By repeated application of product rule:

L.H.S.   =

=

=

=

=

=

= R.H.S  Hence proved.

(ii) By Logarithmic differentiation:

Let

Putting , we get

  Hence proved.