y=(e^(8x))/(2x^(2)-3)
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Originally Posted by wallyg y=(e^(8x))/(2x^(2)-3) Hello and welcome to MathHelpForum. You should understand that this is not a homework service nor is it a tutorial service. Please either post some of your own work on these problems or explain what you do not understand about the question.
This is a job for the quotient rule. If $\displaystyle \displaystyle y=\frac{u}{v} \implies y'=\frac{vu'-uv'}{v^2}$ Also $\displaystyle \displaystyle y=e^{8x} \implies y'=8e^{8x}$
Would the answer be (2x-3)*(8e^(8x)-e^(8x)*2)/(2x^(2)-3)^(2) ?
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 $\displaystyle \displaystyle \frac{e^{8x}}{2x^2-3}$ then $\displaystyle \displaystyle \left(\frac{e^{8x}}{2x^2-3}\right)' = \frac{(2x^2-3)(e^{8x})'-e^{8x}(2x^2-3)'}{(2x^2-3)^2}$
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