use the quotient rule to differentiate the function

k(x)= (e^6x)/(x^3+8)

I have got as far as

k '(x) =(x^3+8)(6e^6x)-(e^6x)(3x^2)/(x^3+8)^2

But im unsure how to take it any further would anyone be able to help us?

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- Mar 8th 2012, 03:38 AMOrlandoAre these the correct derivatives for these functions?
use the quotient rule to differentiate the function

k(x)= (e^6x)/(x^3+8)

I have got as far as

k '(x) =(x^3+8)(6e^6x)-(e^6x)(3x^2)/(x^3+8)^2

But im unsure how to take it any further would anyone be able to help us? - Mar 8th 2012, 11:42 AMSironre: Are these the correct derivatives for these functions?
Your answer is correct! The only thing you can do is put a factor $\displaystyle 3e^{6x}$ outside in the numerator.

- Mar 8th 2012, 01:44 PMOrlandore: Are these the correct derivatives for these functions?
so would that make it

k'(x)= e^6x(6x^3+48-3x^2)/(x^3+8)^2 - Mar 8th 2012, 09:23 PMx3bnmre: Are these the correct derivatives for these functions?
What Siron meant is this:

$\displaystyle \frac{dk}{dx} = \frac{3e^{6x}(2x^{3} + 16 - x^{2})}{(x^{3} + 8)^{2}}$

You just factor $\displaystyle 3e^{6x}$ using simple algebra.

You can also find differentiation of $\displaystyle k(x)= \frac{e^{6x}}{x^3+8}$ in the following link:

differentiate (e^(6x))/(x^3+8) - Wolfram|Alpha