Continuity and Differentiability

55. Kindly go through the solution

54. Kindly go through the solution

53. Kindly go through the solution

52. Kindly go through the solution

51. Given, sin²x + cos²y = 1. 

Differentiating w r t 'x' we get,

$\frac{d}{dx}$ (sin²x + cos²y) $\frac{d}{dx}1.$

$\Rightarrow \frac{d}{dx}si{n}^{2}x+\frac{d}{dx}co{s}^{2}y=0$

$\Rightarrow 2sinx\frac{dsinx}{dx}+2cosy\frac{dcosy}{dx}=0$

= 2sin x cos x + 2 cos y   (- sin y) $\frac{dy}{dx}=0$

= sin 2x- sin 2y $\frac{dy}{dx}$ = 0

50. Given, sin²y + cos xy = p

Differentiating w r t 'x' we get,

$\frac{d}{dx}\left(si{n}^{2}y+cosxy\right)=\frac{d}{dx}$

= $\frac{d}{dx}si{n}^{2}y+\frac{d}{dx}cosxy=0$

$\Rightarrow 2siny\frac{d}{dx}(siny)+(-sinxy)\frac{d}{dx}(xy)=0.$

$\Rightarrow 2sinycosy\frac{dy}{dx}-sinxy\left[x\frac{dy}{dx}+y\right]=0$

$\Rightarrow 2sin2y\frac{dy}{dx}-xsinxy\frac{dy}{dx}-ysinxy=0$

{Qsin 2x = 2sin x cos x}

$\Rightarrow \frac{dy}{dx}[sin2y-xsinxy]=ysinxy.$

$\Rightarrow \frac{dxy}{dx}=\frac{ysinxy}{sin2y-xsinxy}.$

49. Given,  x³ + x²y + xy² + y³ = 81.

Differentiating w r t 'x' we get,

$\frac{d}{dx}\left(x^3+x^{2}y+x{y}^2+y^3\right)$ = $\frac{d(81)}{dx}$

$\Rightarrow \frac{d{x}^3}{dx}+\frac{d}{dx}{x}^{2}y+\frac{d}{dx}x{y}^2+\frac{d}{dx}{y}^3=0$

$\Rightarrow 3x^2+x^2\frac{dy}{dx}+y\frac{d{x}^2}{dx}+\frac{xd{y}^2}{dx}+y^2\frac{dx}{dx}+3y^2\frac{dy}{dx}=0$

$\Rightarrow 3x^2+x^2\frac{dy}{dx}+2xy+2xy\frac{dy}{dx}+y^2+3y^2\frac{dy}{dx}=0.$

$\Rightarrow \left(x^2+2xy+3y^2\right)\frac{dy}{dx}$= - (3x² + 2xy + y²)

$\frac{dy}{dx}=\frac{-\left(3x^2+2xy+y^2\right)}{\left(x^2+2xy+3y^2\right).}$

48. Given,  x² + xy + y² = 100.

Differentiating w r t 'x' we get,

$\frac{d}{dx}\left(x^2+xy+y^2\right)=\frac{d}{dx}(100)$

$\Rightarrow 2x+x\frac{dy}{dx}+y\frac{dx}{dx}+2y\frac{dy}{dx}=0.$

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

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

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

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

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

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

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

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