Derivative Chain Rule

Question

I've been taking the Calculus 1 course on Sophia and am struggling to understand a particular question focused on derivatives and the chain rule. This isn't a test question or anything, and I've already gotten it wrong (can't retake missed questions). I'd like to understand what I've gotten wrong.

The question is:

Find the derivative of $f(x) = 4x^{3}13^{-2x}$.

What I've figured out:

Apply product rule

\begin{align} D[4x^{3} \cdot 13^{-2x}] &= D[4x^3] \cdot 13^{-2x} + 4x^3 \cdot D[13^{-2x}] \\ &= 12x^2 \cdot 13^{-2x} + 4x^3 \cdot D[13^{-2x}] \end{align}

Apply the chain rule

\begin{align} D[a^u] &= (a^u \cdot \ln a) \cdot u' \\ D[13^{-2x}] &= 13^{-2x} \ln(13) \cdot (-2) \end{align}

Adding it together

$$ D[4x^{3} \cdot 13^{-2x}] = 12x^2 \cdot 13^{-2x} + 4x^3 \cdot 13^{-2x} \ln(13) \cdot (-2) $$

The answers I came up with did not match any of the available selections, and I picked one out of frustration (which was incorrect of course).

The correct answer from Sophia

$$ 12x^2 \cdot 13^{-2x} - 8x^3 \cdot 13^{-2x} \ln(13) $$

My question is, where does the $-8x$ come from? Or rather, why is $4x^3$ the only value that is multiplied by $-2$? I've tried using symoblab to break down the answer but it tries to solve the entire equation rather than leaving it as $4x^3 13^{-2x}$.

My goal isn't just to pass the course but to actually understand everything as well but I'm struggling with this one. Any help would be greatly appreciated.

Answer 1

Too long for comment, so consider it an answer, I think You are confused only because you didn't followed your steps by separating them using parantheses

You seem to think that the final $-2$ should be multiplied by Complete expression, which is incorrect

Let's see it again

Apply product rule

$ =( D[4x^3]\cdot 13^{-2x})\ \ +\ \ (4x^3 \cdot D[13^{-2x}])$

$ = (12x^2 \cdot 13^{-2x})\ +\ (4x^3 \cdot D[13^{-2x}])$

Apply the chain rule

$ D[a^u] = (a^u \cdot lna) \cdot u'$

$ = D[13^{-2x}]$

$ = (13^{-2x}ln(13) \cdot -2)$

Clubbing it together

$= (12x^2 \cdot 13^{-2x})+(4x^3 \cdot (13^{-2x}ln(13) \cdot -2))$

$= (12x^2 \cdot 13^{-2x})+(-8x^3 \cdot 13^{-2x}ln(13))$

Which is the correct answer.

Answer 2

Wecolme to MSE: For a simplification of expression$$f(x)=4\frac{x^3}{13^{2x}} \\f'(x)=4.(\frac{x^3}{13^{2x}})'\\=4(\frac{3x^2.{13}^{2x}-\ln(13).{13}^{2x}.(2x)'.x^3}{({13}^{2x})^2})\\= 4(\frac{3x^2.{13}^{2x}-\ln(13).{13}^{2x}.\overbrace{2}^{this} .x^3}{({13}^{2x})^2})$$