use logarithmic differentiation to find the derivative of the function y = (sin x)^(ln x)

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sciencesolve | Teacher | (Level 3) Educator Emeritus

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You need to take logarithms both sides such that:

`ln y = ln((sin x)^(ln x)) `

Using logarithmic identites yields:

`ln y = ln x*ln sin x `

You need to differentiate both sides with respect to x such that:

`(1/y)*y' = (ln x*ln sin x)'`

You should use product rule to the right side such that:

`(1/y)*y' = (ln x)'*ln sin x + (ln x)*(ln sin x)'`

`(1/y)*y' = (1/x)*ln sin x + (ln x)*(1/sin x)*(sin x)'`

`(1/y)*y' = (1/x)*ln sin x + (ln x)*(cos x/sin x)`

`(1/y)*y' = (1/x)*ln sin x + (ln x)*(cot x)`

`y' = y*((1/x)*ln sin x + (ln x)*(cot x))`

Substituting `(sin x)^(ln x) `  for y yields:

`y' = ((sin x)^(ln x))*((1/x)*ln sin x + (ln x)*(cot x)))`

Hence, differentiating the given function, using logarithmic differentiation, yields `y' = ((sin x)^(ln x))*((1/x)*ln sin x + (ln x)*(cot x))).`

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