Given y=x^(cosx), what is dy/dx?

giorgiana1976 | Student

The original form of the given function cannot be differentiated. Therefore, first, we'll have to take natural logarithms both sides:

ln y = ln [x^(cos x)]

Now, we'll apply power property of logarithms:

ln y = cos x*ln x

We'll differentiate both sides, using the product rule to the right side:

y'/y = (cos x)'*ln x + cos x*(ln x)'

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

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

But y = [x^(cos x)]

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

Therefore, dy/dx = [x^(cos x)]*[-sin x*ln x + (cos x)/x].

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