What is f'(x) if f(x)=x^(sin x)?

### 2 Answers | Add Yours

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].**

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**

### Join to answer this question

Join a community of thousands of dedicated teachers and students.

Join eNotes