# What is the result when differentiate f(x)=sin(x^3+square rootx*ln(3x+sin 2x))?

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### 1 Answer

Since we notice that the given functin is a product of composed functions, we'll differentiate with respect to x, using product rule and chain rule.

We'll apply first the product rule:

f'(x)=[sin(x^3 + sqrtx)]'*ln(3x+sin 2x) + [sin(x^3+square rootx)]*[ln(3x+sin 2x)]'

Now, w'ell apply chain rule:

f'(x) = [cos(x^3 + sqrtx)]*(x^3 + sqrtx)'*ln(3x+sin 2x) + [sin(x^3+square rootx)]*[(3 + 2cos 2x)/(3x+sin 2x)]

f'(x) = (3x^2 + 1/2sqrtx)*[cos(x^3 + sqrtx)]*ln(3x+sin 2x) + [sin(x^3+square rootx)]*[(3 + 2cos 2x)/(3x+sin 2x)]

The result of differentiation is:

**f'(x) = (3x^2 + 1/2sqrtx)*[cos(x^3 + sqrtx)]*ln(3x+sin 2x) + [sin(x^3+square rootx)]*[(3 + 2cos 2x)/(3x+sin 2x)]**