Maths

47. Given, xy + y² = tan x + y  Differentiating w r t x we get,

$\frac{d}{dx}\left(xy+y^2\right)=\frac{d}{dx}(tanx+y)$

$\Rightarrow x\frac{dy}{dx}+y\frac{dx}{dx}+\frac{d{y}^2}{dx}=d{x}^{2}x+\frac{dy}{dx}$

$\Rightarrow x\frac{dy}{dx}+2y\frac{dy}{dx}-\frac{dy}{dx}=se{n}^{2}x-y$

$\Rightarrow (x+2y-1)\frac{dy}{dx}=se{c}^{2}x-y$

$\Rightarrow \frac{dy}{dx}=\frac{si{n}^{2}x-y}{x+2y-1.}$

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. L.H.S. = cos 4x.

= cos 2 (2x)

= 1 – 2 Sin² (2x) [ cos 2x = 1 – 2 Sin²x]

= 1 – 2 [2 sin xcosx]² [ sin 2x = 2 sin xcos x]

= 1 – 2 [4 sin²xcos²x]

= 1 – 8 sin²xcos²x

= R.H.S.

40. $$L.H.S. = tan 4x

We know that,

$tan2=\frac{2tan}{1-ta{n}^2}.$ , we can write

$L.H.S=tan2(2x)=\frac{2tan2x}{1-ta{n}^{2}2x.}$

$=\frac{2\left(\frac{2tanx}{1-ta{n}^{2}x}\right)}{1-{\left(\frac{2tanx}{1-ta{n}^{2}x}\right)}^2}$

$=\frac{\frac{4tanx}{\left(1-ta{n}^{2}x\right)}}{\frac{{\left(1-ta{n}^{2}x\right)}^2-4ta{n}^{2}x}{{\left(1-ta{n}^{2}x\right)}^2}}$

$=\frac{4tanx*\left(1-ta{n}^{2}x\right)}{1+ta{n}^{4}x-2ta{n}^{2}x-4ta{n}^{2}x}$

$=\frac{4tanx\left(1-ta{n}^{2}x\right)}{1-6ta{n}^{2}x+ta{n}^{4}x}$

= R.H.S.

39. L.H.S. = cot x cot 2x – cot 2x cot 3x – cot 3x cot x.

= cot x cot 2x – cot 3x (cot 2x + cot x)

= cot x cot 2x – (cot 2x + cot x) [cot (2x + x)]

We know that,

$cot(A+B)=\frac{co\mathrm{tA}co\mathrm{tB}-1}{co\mathrm{tA}+co\mathrm{tB}}$ we can write

$L.H.S=cotxcot2x-(cot2x+cotx)\left[\frac{cot2x·cotx-1}{cot2x+cotx}\right]$

= cot x cot 2x – cot 2x cot x + 1

= R.H.S.