Continuity and Differentiability

42. The given f x v is

f(x) = |x- 1|, x ε R

For a differentiable f x v f at x = c,

$\underset{h\to 0^{-}}{lim}\frac{f(c+h)-f(c)}{h}$ and $\underset{h\to 0^{+}}{lim}\frac{f(c+h)-f(c)}{h}$ are finite & equal.

So, at x = 1. f(1) = |1 - 1| = 0.

Now,

L*H*L* = $\underset{h\to 0^{-}}{lim}\frac{f(1+h-f(1)}{h}$

= $\underset{h\to 0^{-}}{lim}\frac{|1+h-1|-0.}{h}=\underset{h\to 0^{-}}{lim}-\frac{h}{h}$ $\left\{\begin{array}{l}\therefore h<0\\ \left|h\right|=-h\end{array}\right\}$

$=\underset{h\to {\sigma }^{-}}{lim}(-1)$

R*H*L = $\underset{h\to 0^{+}}{lim}\frac{f(1+h)-f(1)}{h}$ = - 1.

$=\underset{h\to 0^{+}}{lim}\frac{(1+h-1)-0}{h}=\underset{h\to 0^{+}}{lim}\frac{h}{h}=\underset{h\to 0^{+}}{lim}1$ $\left\{\begin{array}{cc}\hfill ?f_{n}h>0\hfill & \hfill \left|h\right|=h\hfill \end{array}\right\}$

= 1

Hence, L*H*S ¹ R*H*L*

So, f is not differentiable at x = 2.

41. Kindly consider the following

40. Kindly go through the solution

39. Let f (x) = cos (x³) sin² (x⁵).

f' (x) = cos (x³) $\frac{d}{dx}$  sin² (x⁵) + sin² (x⁵) $\frac{d}{dx}$ cos (x³)

= cos (x³) 2sin (x⁵) $\frac{d}{dx}$  sin (x⁵) + sin² (x⁵) [sin (x³)] $\frac{d}{dx}$ x³.

= 2 cos (x³) sin (x⁵). cos (x⁵) $\frac{d}{dx}$   (x⁵) - sin² (x⁵) sin (x³). 3x²

= 2. cos (x³) sin (x⁵) cos (x⁵). 5 - 3x²sin² (x⁵) sin (x³)

= x² sin (x⁵). [2x² cos (x³) cos (x⁵) - 3 sin (x⁵) sin x³].

38. Let f(x) = $\frac{sin(ax+b)}{cos(cx+d)}.$

f'(x) = $\frac{cos(cx+d)\frac{d}{dx}sin(ax+b)-sin(ax+b)\frac{d}{dx}cos(cx+d)}{co{s}^2(cx+d).}$

= $\frac{cos(cx+d)cos(ax+b)\frac{d}{dx}(ax+b)+sin(ax+b)sin(cx+d)\frac{d(x+d)}{dx}}{co{s}^2(cx+d)}$

= $=\frac{cos(cx+d)cos(ax+b)·a+sin(ax+b)sin(cx+d)·c}{co{s}^2(cx+d)}$

37. Kindly go through the solution

36. Let f (x) = sin (ax + b)

f' (x) = $\frac{d}{dx}$ sin (ax + b)

= cos (ax + b) $\frac{d}{dx}$  (ax + b)

= a cos (ax + b).

35. Let f (x) = cos (sin x).

f' (x) $\frac{d}{dx}$ cos (sin x)

= - sin (sin x) $\frac{d}{dx}$ sin x

= - sin (sin x) cos x.

34. Let f (x) = sin (x2 + 5)

Differentiating w. r t. x we get,

f'¢ (x) = $\frac{d}{dx}$sin (x² + 5)

= cos (x² + 5)  $\frac{d}{dx}$ = cos (x² + 5) 

= cos (x² + 5) [2x].

= 2x cos (x² + 5).