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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)?

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    $\begingroup$ You have $-16x^2$ where it appears to me you need $-8x^2$. ${}\qquad{}$ $\endgroup$ – Michael Hardy Oct 13 '15 at 16:24
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    $\begingroup$ After checking the derivatives, the coefficient of $x^2$ must be $35-16-8+70=81$. $\endgroup$ – Yves Daoust Oct 13 '15 at 16:30
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    $\begingroup$ Why use the product rule? ISTM you'd be better off just expanding everything and doing a term-by-term power rule. $\endgroup$ – Kevin Oct 13 '15 at 19:52
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\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.

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  • $\begingroup$ @MichaelHardy I had intentionally put the wrong calculation to show the OP his mistake. Anyways he has realised it now. $\endgroup$ – SchrodingersCat Oct 13 '15 at 16:28
  • $\begingroup$ thanks to you and Michael both! $\endgroup$ – Analytic Lunatic Oct 13 '15 at 16:45
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$4x^2 \cdot -2 = -8x^2$. You have written $-16x^2$. Purely an arithmetic mistake. Rest looks good to me.

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

So you're $8$ off..

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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 $$f'(x)=80 x^3+81 x^2+66 x+50$$

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  • $\begingroup$ Isn't this what OP computed? Looks the same to me. There is a multiplication error in the next line though. $\endgroup$ – Alfred Yerger Oct 13 '15 at 16:18
  • $\begingroup$ i will expand it $\endgroup$ – Dr. Sonnhard Graubner Oct 13 '15 at 16:19

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