# Derivative of Logarithmic Function

• Oct 1st 2009, 03:38 PM
Maziana
Derivative of Logarithmic Function
How do I derive y = ln(xe^x^3)? Do I do it like this:

Use the formula:
1/(xe^x^3)

Multiply it times the derivative, found using the product rule:
1/(xe^x^3) * (xe^x^3) + e^x^3

Simplify:
((xe^x^3) + e^x^3) / (xe^x^3)

Divide out e^x^3:
(x + 1)/ x

Sorry; I don't know how to make the calculations look good when typing. So, is that right? I have the feeling I did something wrong, so that's why I'm asking. Thanks!
• Oct 1st 2009, 03:43 PM
skeeter
Quote:

Originally Posted by Maziana
How do I derive y = ln(xe^x^3)?

use the laws of logs first ...

$y = \ln(x \cdot e^x \cdot x^3)$

$y = \ln(x) + \ln(e^x) + \ln(x^3)$

$y = \ln(x) + x + 3\ln(x)$

$y = 4\ln(x) + x$

$y' = \frac{4}{x} + 1$
• Oct 1st 2009, 06:07 PM
Maziana
I think I wasn't clear because of my bad formatting...

the problem is not y = ln((x) (e^x) (x^3))

it is y = ((x) (e^x^3)), so e is raised to the xth power, which is then raised to the 3rd power.

Unless you mean I can somehow break it down like that...?
• Oct 1st 2009, 06:12 PM
skeeter
Quote:

Originally Posted by Maziana
I think I wasn't clear because of my bad formatting...

the problem is not y = ln((x) (e^x) (x^3))

it is y = ((x) (e^x^3)), ??? so e is raised to the xth power, which is then raised to the 3rd power.

is it

$y = \ln(x \cdot e^{x^3})$

or is it just

$y = x \cdot e^{x^3}$

???
• Oct 1st 2009, 06:18 PM
Maziana
yeah, the first one... sorry!
• Oct 1st 2009, 06:21 PM
skeeter
$y = \ln(x \cdot e^{x^3})$

once again, use the laws of logs first ...

$y = \ln{x} + \ln(e^{x^3})$

$y = \ln{x} + x^3$

$y' = \frac{1}{x} + 3x^2$
• Oct 1st 2009, 06:27 PM
Soroban
Hello, Maziana!

Quote:

Differentiate: . $y \:=\: \ln\left(xe^{x^3}\right)$
I'll do it two ways . . .

[1] Chain Rule and Product Rule.

. . $y \;=\;\ln\left(x\cdot e^{x^3}\right)$

. . $\frac{dy}{dx} \;=\;\frac{1}{x\cdot e^{x^3}}\cdot\left(x\cdot3x^2\cdot e^{x^3} + 1\cdot e^{x^3}\right) \;=\;\frac{1}{x\cdot e^{x^3}}\cdot e^{x^3}\left(3x^3+1\right)$ . $= \;\frac{3x^3+1}{x}$

[2] Simplify first.

. . $y \;=\;\ln\left(x\cdot e^{x^3}\right) \;=\;\ln(x) + \ln\left(e^{x^3}\right) \;=\;\ln(x) + x^3\cdot\ln(e) \;=\;\ln(x) + x^3$

. . $\frac{dy}{dx} \;=\;\frac{1}{x} + 3x^2 \;=\;\frac{1+3x^3}{x}$