I have the equation: $$f(x) = (5x^2 - 17)e^{-0.5x}$$

I have differentiated it to this so far:

$$f'(x)=10xe^{-0.5x} + (5x^2 -17)(-0.5e^{-0.5x})$$

This is using the chain rule and product rule of differentiation.

I'm having trouble simplifying as there are two common factors: $e^{-0.5x}$ and $-0.5^{-0.5x}$ Normally if there was only $e^{-0.5x}$ it would just simplify to:

$$e^{-0.5x} ( 5x^2 +10x -17) $$

But the other factor is confusing me, so how can I simplify this differentiated equation?

  • $\begingroup$ please use Latex $\endgroup$ – Alex May 20 '14 at 11:16

Your first derivative is slightly incorrect (or you missed a bracket in your question), it should be:


Thus we have the common factor of $e^{-0.5x}$, allowing us to simplify and write:



You do have a common factor, $e^{-0.5x}$ in the following:

$$f'(x)=10x{\pmb{ e^{-0.5x}}} + (5x^2 -17)(-0.5{\pmb{ e^{-0.5x}}})$$

$$f'(x) = e^{-0.5x}\Big(10x -0.5(5x^2-17)\Big)$$

Can you take it from here to simplify the second factor?

If we factor out $-0.5$, that leaves us with $$-0.5e^{-0.5x}\Big(5x^2 -20x -17\Big)$$

  • $\begingroup$ Thanks for the answer, how did the 10x turn into 20x? $\endgroup$ – joe May 20 '14 at 23:18
  • $\begingroup$ Note that I factored out -0.5: $10x = \dfrac{-20x}{-2}= -0.5(-20 x)$ $\endgroup$ – Namaste May 21 '14 at 0:27
  • $\begingroup$ oh right of course, thanks for clarifying that $\endgroup$ – joe May 21 '14 at 1:41

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