Question

y=(sinx-cosx)^sinx-cosx,Find dy/dx for the given function y wherever defined

Answer 1

Given, $\bf{y=(sinx-cosx)^{(sinx-cosx)}}$

Let (sinx - cosx) = P

differentiate with respect to x

d(sinx - cosx)/dx = dP/dx

cosx + sinx = dP/dx ..........(1)

now, y = P^P = e^{PlogP}

differentiate with respect to x,

dy/dx = e^{PlogP} × d(P.logP)/dx

= P^P × [P × 1/P + logP ]dP/dx
= P^P(1 + logP) × dP/dx

from equation (1),

dy/dx = (sinx - cosx)^(sinx - cosx){1 + log(sinx-cosx)} × (sinx + cosx)

Answer 2

In the attachment I have answered this problem.                 I have applied logarithmic differentiation to find the derivative of the given function.                   See the attachment for detailed solution.