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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!
use the laws of logs first ...Quote:
Originally Posted by Maziana
$\displaystyle y = \ln(x \cdot e^x \cdot x^3)$
$\displaystyle y = \ln(x) + \ln(e^x) + \ln(x^3)$
$\displaystyle y = \ln(x) + x + 3\ln(x)$
$\displaystyle y = 4\ln(x) + x$
$\displaystyle y' = \frac{4}{x} + 1$
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...?
is itQuote:
Originally Posted by Maziana
$\displaystyle y = \ln(x \cdot e^{x^3})$
or is it just
$\displaystyle y = x \cdot e^{x^3}$
???
yeah, the first one... sorry!
$\displaystyle y = \ln(x \cdot e^{x^3})$
once again, use the laws of logs first ...
$\displaystyle y = \ln{x} + \ln(e^{x^3})$
$\displaystyle y = \ln{x} + x^3$
$\displaystyle y' = \frac{1}{x} + 3x^2$
Hello, Maziana!
I'll do it two ways . . .Quote:
Differentiate: .$\displaystyle y \:=\: \ln\left(xe^{x^3}\right)$
[1] Chain Rule and Product Rule.
. . $\displaystyle y \;=\;\ln\left(x\cdot e^{x^3}\right)$
. . $\displaystyle \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) $ .$\displaystyle = \;\frac{3x^3+1}{x}$
[2] Simplify first.
. . $\displaystyle y \;=\;\ln\left(x\cdot e^{x^3}\right) \;=\;\ln(x) + \ln\left(e^{x^3}\right) \;=\;\ln(x) + x^3\cdot\ln(e) \;=\;\ln(x) + x^3$
. . $\displaystyle \frac{dy}{dx} \;=\;\frac{1}{x} + 3x^2 \;=\;\frac{1+3x^3}{x} $
