# Are these the correct derivatives for these functions?

• Mar 8th 2012, 03:38 AM
Orlando
Are 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 AM
Siron
re: 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 PM
Orlando
re: 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 PM
x3bnm
re: Are these the correct derivatives for these functions?
Quote:

Originally Posted by Orlando
so would that make it

k'(x)= e^6x(6x^3+48-3x^2)/(x^3+8)^2

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