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Use chain rule of differentiation and find derivative of f(x)=(x^3+4)^4?
We'll use the chain rule of differentiation to find the derivative of a function, that is the result of composition of 2 or more functions.
We'll apply chain rule to function f(x) = (x^2+sinx)^2
u(x) = x^2+sinx and v(u) = u^2
f(x) = (vou)(x) = v(u(x)) = v(x^2+sinx) = (x^2+sinx)^2
We'll differentiate f(x) and we'll get:
f'(x) = v'(u(x))*u'(x)
First, we'll differentiate v with respect to u:
v'(u) = 2u^(2-1) = 2u
Second, we'll differentiate u with respect to x:
u'(x) = (x^2+sin x)' = 2x + cos x
f'(x) = 2u*(2x + cos x)
We'll substitute u and we'll get:
The derivative of f(x) is: f'(x) = 2(x^2+sinx)*(2x + cos x)
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