Question

y= (sinx)^cos x. find out ​

Answer 1

Answer:

dy/dx = sinx^cosx(- sinxln(sinx) + cosx cotx)

Step-by-step explanation:

Take logarithm of both sides.

lny=ln(sinx)^cosx

lny=cosxln(sinx)

Use the implicit differentiation as well as the product and chain rules to differentiate.

d/dx(lnsinx) = 1/sinx(cosx)

= cosx/sinx

= cotx

now,

1/y(dy/dx) = - sinx(ln(sinx)) + cosx cotx

dy/dx = y(- sinxln(sinx) + cosx cotx)

= sinx^cosx(- sinxln(sinx) + cosx cotx)

hope u understand