Question

Differentiate x^y=e^x-y​

Answer 1

Step-by-step explanation:

applying log on both sides of given equation

log(x^y)=log(e^(x-y))

ylog(x)=(x-y)log(e)

ylog(x)=(x-y)(1)

ylog(x)=x-y

differentiate with respect to x

using UV rule i.e. (d(UV)/dx)=v(du/dx)+u(dv/dx)

and chain rule

y(d(logx)/dx)+(logx)(dy/dx)=(dx/dx)-(dy/dx)

y(1/x)+(logx)(dy/dx)=1-(dy/dx)

(y/x)+(logx)(dy/dx)=1-(dy/dx)

(logx)(dy/dx)+(dy/dx)=1-(y/x)

(dy/dx)(logx+1)=(x-y)/x

(dy/dx)=[(x-y)/x][1/(logx+1)]

(dy/dx)=(x-y)/x(logx+1)

(dy/dx)=(x-y)/(xlogx+x)