Question

If y^y=xsiny,then dy/dx is,​

Answer 1

Step-by-step explanation:

We use what's called the product rule for differentiating equation in the form of u.v;

y=u.v

dydx=u∗dvdx+v∗dudx

For this case we are also going to use the fact that;

ddxf(y)=f′(y)dydx

This is true by chain rule as you can see below;

u=f(y)

dudx=dudy∗dydx

So;

y=x∗sin(y)

We let u=x and v=sin(y)

We differentiate both sides with respect to X;

dydx=x∗cos(y)dydx+sin(y)

We have to make dydx the subject so we bring them on one side;

dydx−x∗cos(y)dydx=sin(y)

Factorising gives;

dydx(1−xcos(y))=sin(y)

dydx=sin(y)1−xcos(y)

Answer 2

Answer:

Step-by-step explanation:

ln(y^y)= ln(xsiny)

ylny= ln(x) +ln(siny)

d(ylny)/dx= d(lnx)/dx +d(ln(siny))/dx

lny. dy/dx+ dy/dx= 1/x+ coty.dy/dx

dy.(lny+1-coty)/dx= 1/x

dy/dx= 1/{x(1+lny-coty)}