Question

sin2y+cosxy=k,find dy/dx​

Answer 1

Answer:

$\frac{dy}{dx} =\frac{ysin(xy)}{sin2y-xsinxy}$

Step-by-step explanation:

sin²y + cos xy = k

differentiating both side w.r.t.x.

$\frac{d(sin ^{2}+cos xy)}{dx}=\frac{d(k)}{dx}$

hence derivative of constant is zero  

$\frac{d(sin^{2}y)}{dx} +\frac{d(cos xy)}{dx}=0$

calculating derivative of sin²y and cos(xy) sperately

calculating derivative of sin²y

$\frac{d(sin^2 y)}{dx}= \frac{d(sin^{2}y)}{dx}  *\frac{dy}{dy}$

$= \frac{d(sin^{2}y)}{dx} *\frac{dy}{dx}$

$=2sin^{2-1}x.\frac{d(sin y)}{dy} * \frac{dy}{dx}$

=$=2siny.cosy*\frac{dy}{dx}$

calculating derivative of cos(xy)

$\frac{d(cos(xy))}{dx}=-sin(xy)*\frac{d}{dx} (xy)$

$=-sin(xy)*(\frac{d(x)}{dx}.y+\frac{d(y)}{d(x)}.x )$

$=-sin(xy).(1.y+x\frac{dy}{dx} )$

$=-sin(xy)(y+x\frac{dy}{dx} )$

$=-sin(xy).y-sin(xy).x\frac{dy}{dx}$

$=-ysin(xy)-sin(xy).x\frac{dy}{dx}$

now,

$\frac{d(sin^{2}y)}{dx}+\frac{d(cos(xy))}{dx}=0$

putting values

$2siny.cosy.\frac{dy}{dx} +(-ysin(xy)-xsin(xy)\frac{dy}{dx} )=0$

$2siny.cosy.\frac{dy}{dx}-ysin(xy)-xsin(xy)\frac{dy}{dx}=0$

$2siny cosy\frac{dy}{dx}-xsin(xy)\frac{dy}{dx}=y sin(xy)$

$\frac{dy}{dx}(2sinycosy-xsin(xy))=ysin(xy)$

$\frac{dy}{dx}=\frac{ysin(xy)}{2sinycosy-xsinxy}$

$\frac{dy}{dx}=\frac{ysin(xy)}{sin2y-xsinxy}$