Differentiate siny=xy3+ylnx?

2 Answers
Aug 30, 2017

dydx=xy3+yxcosy3x2y2xlnx

Explanation:

siny=xy3+ylnx
Differentiating both sides with respect to x we get

cosydydx=y3+x×3y2dydx+yx+lnx×dydx
xcosydydx=xy3+3x2y2dydx+y+xlnxdydx
dydx=xy3+yxcosy3x2y2xlnx

Aug 30, 2017

dydx=xy3+y(xcosy3x2y2xlnx)

Explanation:

differentiate implicitly with respect to x

differentiate xy3 and ylnx using the product rule

cosy.dydx=(x.3y2dydx+y3)+(yx+lnx.dydx)

dydx(cosy3xy2lnx)=y3+yx

dydx=y3+yxcosy3xy2lnx

rAeedydx=xy3+yxcosy3x2y2xlnx