Class 12th

57. Kindly go through the solution

56. Kindly go through the solution

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)

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