Find f' (x) of a function: f(x) = 2^(4x) + 4x^2

Question

I have to find f'(x) for the function f(x)=2^(4x) + 4x^2

I know the answer is supposed to be 2^(4x+2) (ln2+8x), but I'm not sure how they got this. Any help would be much appreciated!

Answer 1

The two following rules are applicable in this case:

Additionally, you will need to apply the chain rule.

For 2^4x use the exponential rule where a = 2 PLUS use the chain rule since the exponent is more complicated than just "x". So the derivative is [ln(2)](2^4x)(4) = [ln(2)](2^4x)(2²) = ln(2)(2^4x+2)

For 4x², use the power rule so the derivative is 8x

Combine these together to get f '(x) = ln(2)(2^4x+2) + 8x

Note that your answer has parenthesis in the wrong place.