What is f'(x) if f(x)=x^(sin x)?
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The function f(x) = x^(sin x)
Let y = f(x) = x^(sin x)
Take the natural log of both the sides
ln y = ln [ x^(sin x)]
=> ln y = sin x * ln x
Differentiate both the sides with respect to x
=> (1/y)(dy/dx) = (sin x)/x + cos x * ln x
=> dy/dx = y*[sin x + (cos x)(ln x)*x]/x
=> dy/dx = [(x^(sin x))*(sin x) + x^(sin x)(cos x)(ln x)*x]/x
=> dy/dx = [(x^(sin x))*(sin x) + x^(sin x + 1)(cos x)(ln x)]/x
f'(x) = [(x^(sin x))*(sin x) + x^(sin x + 1)(cos x)(ln x)]/x
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First thing, we'll take natural logarithms both sides:
ln f(x) = ln [x^(sin x)]
We'll apply the power rule of logarithms:
ln f(x) = sin x* ln x
We'll differentiate with respect to x both sides:
f'(x)/f(x) = (sin x*ln x)'
We'll apply product rule to the right side:
f'(x)/f(x) = cos x* ln x + (sin x)/x
Now, we'll multiply both sides by f(x):
f'(x) = f(x)*[cos x* ln x + (sin x)/x]
But f(x) = x^(sin x), therefore f'(x) = [x^(sin x)]*[cos x* ln x + (sin x)/x].
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