What is the derivative of $y = (sin x)^(cos x)$ ?

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Answer 1 · 2017-01-04T16:26:53

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

Take the natural logarithm of both sides.

$lny = ln(sinx)^cosx$

Use the rule $loga^n = nloga$ to simplify:

$lny = cosxln(sinx)$

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

$d/dx(lnsinx) = 1/sinx * cosx = cosx/sinx = cotx$

Now to the relation as a whole:

$1/y(dy/dx) = -sinx(ln(sinx)) + cosxcotx$

Solve for $dy/dx$ :

$dy/dx = y(-sinxln(sinx) + cosxcotx)$

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

Hopefully this helps!

What is the derivative of y = (sin x)^(cos x)y = (sin x)^(cos x)?