Finding the derivative for a product of two polynomial functions?

Question

In my problem, I am attempting to find $f'(x)$ when $f(x)=(5x^2-2x+8)(4x^2+7x-3)$. For my work I have:

\begin{align} & \frac{d}{dx} (uv) = u\frac {dv}{dx} + v\frac {du}{dx} \\[8pt] = {} & (5x^2-2x+8)(8x+7)+(4x^2+7x-3)(10x-2) \\[8pt] = {} & 40x^3+35x^2-16x^2-14x+64x+56+40x^3-16x^2+70x^2-14x-30x+6 \\[8pt] = {} & 80x^3 + 73x^2+6x+62 \end{align}

But when I plugged my original equation [$f(x)=(5x^2-2x+8)(4x^2+7x-3)$] into an online derivative calculator to check my answer, it comes out as:

$$=80x^3+81x^2+6x+62\ldots\text{ ?}$$

Can anyone spot where I am going wrong (if I am)?

Answer 1

\begin{align} & \frac{d}{dx} (uv) = u\frac {dv}{dx} + v\frac {du}{dx} \\[8pt] = {} & (5x^2-2x+8)(8x+7)+(4x^2+7x-3)(10x-2) \\[8pt] = {} & 40x^3+35x^2-16x^2-14x+64x+56+40x^3-16x^2+70x^2-14x-30x+6 \\[8pt] = {} & 80x^3 + 73x^2+6x+62 \end{align}

In the penultimate line of calculation, the 8th term from left should be $$4x^2 \times (-2) = -8x^2$$ instead of $(-16x^2)$. Correct that and then the answers will match.

Answer 2

$4x^2 \cdot -2 = -8x^2$. You have written $-16x^2$. Purely an arithmetic mistake. Rest looks good to me.

Answer 3

HINT: after the product rule we get $$f'(x)=(10x-2)(4x^2+7x-3)+(5x^2-2x+8)(8x+7)$$ after expanding we get $$f'(x)=80 x^3+81 x^2+66 x+50$$

Answer 4

When you multiply the second set of parenthesis, you write $4 \cdot -2 = -16$ instead of $-8$.

So you're $8$ off..