# Derivatives of Exponential Functions Question

1. Nov 13, 2006

### TommyLF

The problem is:

y= 6^3x-7
--------
(x^2)-x
Read: Y equals 6 to the 3x minus 7 OVER/Divided by x to the 2nd minus x

Now I thought I had to do a quotient rule there and start by doing:

(x^2-x)(e^(ln6)(3x-7))(3) and then so on from there. I just don't know what to do with that 6 to the 3x-7. Is it something to do with "e"?

2. Nov 13, 2006

### Office_Shredder

Staff Emeritus
$$6^{3x-7} = e^{(3x-7)*ln(6)}$$ You can see this by rewriting it as $$(e^{ln(6)})^{3x-7}$$ although the first form is easier to differentiate. So the derivative, by the chain rule, is $$e^{3x-7)*ln(6)}*ln(6)*3$$ or $$6^{3x-7}*3ln(6)$$

3. Nov 13, 2006

### TommyLF

So would the quotient rule look like:

(x^2-x)(e^(3x-7)(ln 6))(3ln 6) + (6^3x-7)(2x-1)
--------------------------------------------
(x^2-x)^2

By the way, how do I get my question to look like nice and neat without a bunch of ^ like you just did?

4. Nov 13, 2006

### calcnd

Is a mathematical markup language called LaTeX.

5. Nov 14, 2006

### HallsofIvy

Staff Emeritus
More specifically, the derivative of ax is ln(a) ax.

If you click on something like

$$\frac{da^x}{dx}= ln(a) a^x$$
A popup box will show the code.