Okay I know how to go about getting a derivative and 2nd derivatives. However, for this problem I can't figure out how to get to the right answer. this is the function f(x) = x(x-4)^3
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Originally Posted by sgonzalez90 Okay I know how to go about getting a derivative and 2nd derivatives. this is the function f(x) = x(x-4)^3 The first derivative is $\displaystyle f'(x)=(x-4)^3+3x(x-4)^2$. You finish.
Originally Posted by Plato The first derivative is $\displaystyle f'(x)=(x-4)^3+3x(x-4)^2$. You finish. Yes I got the first derivative already. I wasn't able to get the 2nd.
Originally Posted by sgonzalez90 Yes I got the first derivative already. I wasn't able to get the 2nd. Use a combination of the product rule and the chain rule. Hint: $\displaystyle (x-4)^3+3x(x-4)^2 = (x-4)^3 + (3x \times (x-4)^2)$
You cannot find the derivative of $\displaystyle (x-4)^3+3x(x-4)^2$? You are joking, right?
Originally Posted by Plato You cannot find the derivative of $\displaystyle (x-4)^3+3x(x-4)^2$? You are joking, right? I tried this and got a big mess of a problem but this is what i got 3x-12 (x-4)^2 + (6x^2-24x)^2(x-4)
$\displaystyle f'(x)=(x-4)^3+3x(x-4)^2$ $\displaystyle f"(x)=3(x-4)^2+3(x-4)^2+6x(x-4)$ That is it.
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