# Use chain rule of differentiation and find derivative of f(x)=(x^3+4)^4?

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The chain rule of differentiation gives the derivative of f(g(x)) as f'(g(x))*g'(x)

For f(x) = (x^3 + 4)^4

we can take u(x) = x^3 + 4

f(x) = (u(x))^4

f'(x) = 4*(u(x))^3*u'(x)

=> f'(x) = 4*(x^3 + 4)^3 * 3x^2

=> f'(x) = 12x^2*(x^3 +4)^3

**The derivative of (x^3 + 4)^4 = 12x^2*(x^3 +4)^3**

To use the chain rule, we'll specify first that f(x) is the result of composition of 2 functions.

u(x) = x^3 + 4 and v(u) = u^4

f(x) = (vou)(x) = v(u(x)) = v(x^3 + 4) = (x^3 + 4)^4

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) = 4u^(4-1) = 4u^3

Second, we'll differentiate u with respect to x:

u'(x) = (x^3 + 4)' = 3x^2

f'(x) = 4u^3*3x^2

We'll substitute u and we'll get:

**The derivative of f(x) is: f'(x) = 12x^2*( x^3 + 4)^3.**