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I came across the problem on Khan Academy while studying differential calculus:

Consider the function $f(x) = e^{2x}(x^2 + 2x)$.

There are two x-coordinates at which $f'(x) = f''(x)$. What is the sum of these two coordinates?

While finding the derivative of $f(x)$, I got everything reduced down to

$$f'(x) = e^{2x}(2x+2) + 2e^{2x}(x^2 + 2x)$$

Khan Academy says this can be further reduced to $e^{2x}(2x^2 + 6x + 2)$, obviously so I can apply the product rule again to find the second derivative, but I have no idea how they made that happen. Can anyone help me understand their algebra?

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  • $\begingroup$ It's just basic factorizing, isn't it? $\endgroup$
    – GorTeX
    Commented Mar 28, 2015 at 21:51
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    $\begingroup$ HINT: First distribute the $2$ in the second term into the brackets as follows $$2e^{2x}(x^2+2x)=e^{2x}(2x^2+4x)$$ $\endgroup$
    – Mufasa
    Commented Mar 28, 2015 at 21:52

3 Answers 3

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$f'(x) = e^{2x}\left(2x+2+2(x^2+2x)\right) = e^{2x}\left(2x^2+6x+2\right)$. Can you take it from here?

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  • $\begingroup$ Derp. Thank you! Yes, I'm going to apply the product rule to the reduced first derivative, then solve for $f'(x) = f''(x)$. I'll accept your answer in seven minutes. $\endgroup$
    – sent1nel
    Commented Mar 28, 2015 at 21:55
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Just factor out the $e^{2x}$:

$$e^{2x}(2x+2)+2e^{2x}(x^2+2x)=e^{2x}(2x+2+2(x^2+2x))=e^{2x}(2x^2+6x+2)$$

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Facor out as was shown above: $$f'(x) = e^{2x}\left(2x+2+2(x^2+2x)\right) = e^{2x}\left(2x^2+6x+2\right)$$ But to find the second derivative you don't have to do it, because: $$f''=(e^{2x}(2x+2))' + (2e^{2x}(x^2 + 2x))'=2e^{2x}(2x+3)+4e^{2x}(x^2+3x+1)=2e^{2x}(2x^2+8x+5)$$

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