How do you differentiate ${(cos x)}^{sin x}$ $(cos x)^(sin x)$ ?

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Answer 1 · 2016-07-11T05:45:58

$y' = (cos x)^(sin x)(sin x * (-sin x)/(cos x) + cos x * ln(cos x))$

This problem requires us to make use of logarithmic differentiation, as well as the product rule for derivatives.

Let $y = (cosx)^(sinx)$

Taking the natural logarithm of both sides yields

$ln(y) = sin x * ln(cosx)$

Now we have to take a derivative on each side of the equation.

Remembering that

$d/dx[ln(u)] = (u')/(u)$ and $d/dx[f(x)g(x)] = f(x)g'(x)+f'(x)g(x)$

We now get

$(y')/(y) = sin x * (-sin x)/(cos x) + cos x * ln(cos x)$

Isolating our $y'$ -term gives us

$y' = y(sin x * (-sin x)/(cos x) + cos x * ln(cos x))$

Substituting $y$ back into the equation gives us our final result of

$y' = (cos x)^(sin x)(sin x * (-sin x)/(cos x) + cos x * ln(cos x))$

How do you differentiate (cos x)^(sin x)(cosx)sinx (cos x)^(sin x)?