Continuity and Differentiability

46. Given, ax + by2 = cos y.

Differentiating w r t 'x' we get,

$\frac{d}{dx}\left(ax+b{y}^2\right)=\frac{dx}{dx}cosy$

= a + b 2y = - sin y $\frac{dy}{dx}$ + sin y $\frac{dy}{dx}$ = -a

= $\frac{dy}{dx}=\frac{-9}{2by+siny}.$

= 2by $\frac{dy}{dx}$

45. Given, 2x + 3y = sin y.

Differentiating w r t x. we get,

$\frac{d}{dx}(2x+3y)=\frac{d}{dx}siny$

$\Rightarrow 2+3\frac{dy}{dx}=cosy\frac{dy}{dx}$

=cos y $\frac{dy}{dx}-3\frac{dy}{dx}=2$

$\Rightarrow \frac{dy}{dx}(cosy-3)=2$

= $\frac{dy}{dx}=\frac{2}{cosy-3}$

44. Kindly go through the solution

43. The given f x ⁿ is

f(x) = 0 < $\left|x\right|$ < 3

At x = 1

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

$=\underset{h\to 0^{-}}{lim}\frac{[1+h]-[1]}{h}$

$\underset{h\to \sigma }{lim}\frac{0-1}{h}$ $\left\{\begin{array}{ccc}\hfill ?h<0\hfill & \hfill ,1+h<1-1\hfill & \hfill So, [1+h]=0\hfill \end{array}\right\}$

$=\underset{h\to 0}{lim}\frac{-1}{h}=\infty$

Hence lines does not exist

Qf is not differentiable at x = 1

At x = 2

L*H*L = $\underset{h\to 0^{-}}{lim}\frac{f(2+h)-f(2)}{h}$ $\left\{\begin{array}{ccc}\hfill ?h<0\hfill & \hfill 2+h<2\hfill & \hfill 30,[2+h]=1\hfill \end{array}\right\}$

$=\underset{h\to 0^{-}}{lim}\frac{[2+h]-[2].}{h}$

$=\underset{h\to 0^{-}}{lim}\frac{1-2}{h}=\underset{h\to 0^{-}}{lim}\frac{-1}{h}$

Hence, limit does not exist.

Qf is not differentiable at x = 2

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

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

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