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Take the natural log of both sides?

  • Thread starter tony873004
  • Start date

tony873004

Science Advisor
Gold Member
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Find the first derivitive. Simplify if possible (factor).
[tex]
\begin{array}{l}
y = x^{e^x } \\
\\
\ln y = \ln x^{e^x } \\
\\
\ln y = e^x \ln x \\
\end{array}
[/tex]
There's a similar problem in my class notes where it was solved by taking the natural log of both sides. Is this the way to go on this problem? If so, I'm stuck at this point.
 

Pengwuino

Gold Member
4,848
10
Can't you just use the chain rule?
 
tony873004 said:
Find the first derivitive. Simplify if possible (factor).
[tex]
\begin{array}{l}
y = x^{e^x } \\
\\
\ln y = \ln x^{e^x } \\
\\
\ln y = e^x \ln x \\
\end{array}
[/tex]
There's a similar problem in my class notes where it was solved by taking the natural log of both sides. Is this the way to go on this problem? If so, I'm stuck at this point.
You're doing just fine....differentiat both sides to get
[tex]
\frac{1}{y} \frac{dy}{dx} = e^xln(x)+e^x/x
[/tex]

Then go from there....
 

Curious3141

Homework Helper
2,830
85
tony873004 said:
Find the first derivitive. Simplify if possible (factor).
[tex]
\begin{array}{l}
y = x^{e^x } \\
\\
\ln y = \ln x^{e^x } \\
\\
\ln y = e^x \ln x \\
\end{array}
[/tex]
There's a similar problem in my class notes where it was solved by taking the natural log of both sides. Is this the way to go on this problem? If so, I'm stuck at this point.
As Geekster said, you can use implicit differentiation. Personally, I'd just use chain rule.

[tex]y = x^{e^x} = e^{(e^x)(\ln x)}[/tex]

This is of the form [tex]y = e^{f(x)}[/tex] the derivative of which is

[tex]y' = f'(x)e^{f(x)}[/tex]

So
[tex]y' = (\frac{1}{x}e^x + (e^x)(\ln x))(e^{(e^x)(\ln x)})[/tex]

which you can simplify.
 
Last edited:

tony873004

Science Advisor
Gold Member
1,749
141
re: geekster
Thanks. I got that far. Should I just multiply both sides by y? But that would leave me with [tex]
\frac{{dy}}{{dx}} = e^x y\ln x + \frac{1}{x}y
[/tex]
Don't I need to get rid of y on the right side?

re: Curious3141
Thanks.
This was on the test for implicit differentiation. Chain rule was last test, but he didn't say we couldn't use it. Give me a little while to see if I can simplify that, and I'll post what I get.
 
Last edited:
tony873004 said:
re: geekster
Thanks. I got that far. Should I just multiply both sides by y? But that would leave me with [tex]
\frac{{dy}}{{dx}} = e^x y\ln x + \frac{1}{x}y
[/tex]
Don't I need to get rid of y on the right side?
No....you were given y in the problem. Just replace y by [tex]x^e^x[/tex] and you get the same answer that Curious3141 gave. I think it's good to see the same answer can come from many different methods. Although Curious3141's method is more elegant IMO.
 
is this how u do it..?

ya.jpg
 

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