How do you find the derivative of $(ln x) ^ cos x$ ?

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Answer 1 · 2016-07-13T07:47:53

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

For this particular, we'd have to use logarithmic differentiation, which works as follows:

Let $y = (ln x)^(cos x)$

Taking the natural log ( $ln$ ) of both sides yields

$ln y = ln((ln x)^(cos x))$

$ln y = cos x * ln(ln x)$

Since the next step is to take derivatives, the rules we're going to use is

$d/dx[ln u] = (u')/(u)$

Differentiating both sides gives

$(y')/(y) = -sin x * ln(ln x) + cos x * ((1/x) * (1/ln x))/(((cancel(ln x))/1) * (1/cancel(ln x)))$

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

But since we're looking for $y'$ , we simply move $y$ to the right side of our equation to get

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

Substituting $y$ back into the equation yields

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

How do you find the derivative of (ln x) ^ cos x (ln x) ^ cos x?