# Thread: Finding derivative of a Exponential Function

1. ## Finding derivative of a Exponential Function

y=(e^(8x))/(2x^(2)-3)

2. Originally Posted by wallyg
y=(e^(8x))/(2x^(2)-3)
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3. This is a job for the quotient rule.

If $\displaystyle y=\frac{u}{v} \implies y'=\frac{vu'-uv'}{v^2}$

Also $\displaystyle y=e^{8x} \implies y'=8e^{8x}$

4. Would the answer be (2x-3)*(8e^(8x)-e^(8x)*2)/(2x^(2)-3)^(2) ?

5. Originally Posted by wallyg
Would the answer be (2x-3)*(8e^(8x)-e^(8x)*2)/(2x^(2)-3)^(2) ?
I don't think so, given your function is $\displaystyle \frac{e^{8x}}{2x^2-3}$ then

$\displaystyle \left(\frac{e^{8x}}{2x^2-3}\right)' = \frac{(2x^2-3)(e^{8x})'-e^{8x}(2x^2-3)'}{(2x^2-3)^2}$