Thread: Finding x (this is really beyond me i think)???

This set of questions is really perplexing me!!!!

find f '(x) given that

f (x) = 2x^3 + 6x^2 - (cubed root of x)

f (x) = x^2 e^(-3x)

These really phase me a bit, any advice much appreciated.

Originally Posted by bobchiba

This set of questions is really perplexing me!!!!

find f '(x) given that

f (x) = 2x^3 + 6x^2 - (cubed root of x)

f (x) = x^2 e^(-3x)

These really phase me a bit, any advice much appreciated.

What's the problem with the first one? It's just the power rule:
d/dx(ax^n) = a*n*x^{n-1}

So:
f(x) = 2x^3 + 6x^2 - x^{1/3}

f'(x) = 6x + 12x - (1/3)x^{-2/3}

Use the product rule:
h(x) = f(x)*g(x) ==> h'(x) = f'(x)*g(x) + f(x)*g'(x)

So for
f(x) = x^2 e^{-3x}

f'(x) = 2x*e^{-3x} + x^2 * (-3)e^{-3x} = 2xe^{-3x} - 3x^2e^{-3x}

-Dan