Derivatives of Exponential Functions Question

[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"?

 

[Office_Shredder]

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)

 

[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?

 

[calcnd]

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

Is a mathematical markup language called LaTeX.

Here's some more info for this forum's LaTeX engine:

https://www.physicsforums.com/showthread.php?t=8997

 

[HallsofIvy]

More specifically, the derivative of a^x is ln(a) a^x.

If you click on something like

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