NCERT Solutions for Class 12 Maths Exercise 5.5 Chapter 5 Continuity and Differentiability – FREE PDF Download
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NCERT Solutions for Class 12 Maths Chapter 5 Continuity and Differentiability Exercise 5.5 (Ex 5.5)
Differentiate the functions with respect to 
in Exercise 1 to 5.
1. 
……….(i) Taking logs on both sides, we have
log y = 
2. 
= 
……….(i) Taking logs on both sides, we have
3.
……….(i) Taking logs on both sides, we have
[By Product rule]
1y.dydx=cosxlogx.1x−sinxlog(logx)1y.dydx=cosxlogx.1x−sinxlog(logx)
dydx=y[cosxxlogx−sinxlog(logx)]dydx=y[cosxxlogx−sinxlog(logx)]
⇒⇒ dydx=(logx)cosx[cosxxlogx−sinxlog(logx)]dydx=(logx)cosx[cosxxlogx−sinxlog(logx)]
4.
Putting 
and 
……….(i)
Now,
dudxdudx= 
……….(ii)
Again,
……….(iii)
Putting the values from eq. (ii) and (iii) in eq. (i),
5.
…….(i) Taking logs on both sides, we have
logy=2log(x+3)+3log(x+4)+4log(x+5)logy=2log(x+3)+3log(x+4)+4log(x+5)
dydx=(x+3)2(x+4)3(x+5)4(2(x+4)(x+5)+3(x+3)(x+5)+4(x+3)(x+4)(x+3)(x+4)(x+5))dydx=(x+3)2(x+4)3(x+5)4(2(x+4)(x+5)+3(x+3)(x+5)+4(x+3)(x+4)(x+3)(x+4)(x+5))
dydx=(x+3)2(x+4)3(x+5)4(2x2+18x+40+3x2+24x+45+4x2+28x+48(x+3)(x+4)(x+5))dydx=(x+3)2(x+4)3(x+5)4(2×2+18x+40+3×2+24x+45+4×2+28x+48(x+3)(x+4)(x+5))
dydx=(x+3)(x+4)2(x+5)3(9x2+70x+133)dydx=(x+3)(x+4)2(x+5)3(9×2+70x+133)
Differentiate the functions with respect to in Exercise 6 to 11.
6. (x+1x)x+x(1+1x)(x+1x)x+x(1+1x)
Putting 
and x(1+1x)=vx(1+1x)=v
……….(i)
Now 
1u.dudx=x.1(x+1x)(1−1x2)+log(x+1x).11u.dudx=x.1(x+1x)(1−1×2)+log(x+1x).1
= 
……….(ii)
Again v=x(1+1x)v=x(1+1x)
logv=logx(1+1x)=(1+1x)logxlogv=logx(1+1x)=(1+1x)logx
1v.dvdx=(1+1x).1x+logx.(−1x2)1v.dvdx=(1+1x).1x+logx.(−1×2)
dvdx=v[(1+1x).1x+logx.(−1x2)]dvdx=v[(1+1x).1x+logx.(−1×2)]
dvdx=x(1+1x)[1x(1+1x)−1x2logx]dvdx=x(1+1x)[1x(1+1x)−1x2logx] ……….(iii)
Putting the values from eq. (ii) and (iii) in eq. (i),
dydx=(x+1x)x[x2−1x2+1+log(x+1x)]+x(1+1x)[1x(1+1x)−1x2logx]dydx=(x+1x)x[x2−1×2+1+log(x+1x)]+x(1+1x)[1x(1+1x)−1x2logx]
7.
= 
where 
and 
……….(i)
Now
……….(ii)
Again
……….(iii)
Putting the values from eq. (ii) and (iii) in eq. (i),
8.
= where 
and 
……….(i)
Now
1u.dudx=x.1sinxcosx+log(sinx)=xcotx+logsinx1u.dudx=x.1sinxcosx+log(sinx)=xcotx+logsinx
……….(ii)
Again
= 
= 
……….(iii)
Putting the values from eq. (ii) and (iii) in eq. (i),
9.
Putting 
and 
, we get
……….(i)
Now
= 
…..(ii)
Again
= 
……….(iii)
Putting values from eq. (ii) and (iii) in eq. (i),
10.
Putting 
and 
,
we have
……….(i)
Now
= 
……….(ii)
Again
……….(iii)
Putting the values from eq. (ii) and (iii) in eq. (i),
11.
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. 
where 
and 
……….(i)
Now
……….(ii)
Again
……….(iii)
Putting values from eq. (ii) and (iii) in eq. (i),
13.
14.
15.
16. Find the derivative of the function given by and hence
……….(i) 
Putting the value of 
from eq. (i),
8 x 15 = 120
17. Differentiate in three ways mentioned below:
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?
……….(i) (i) 
dydx=5x4−20x3+45x2−52x+11dydx=5×4−20×3+45×2−52x+11 ……….(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.
then show that 
in two ways – first by repeated application of product rule, second by logarithmic differentiation.
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.
