# Logarithmic Differentiation Question

• Jan 1st 2011, 12:42 PM
ashleysmithd
Logarithmic Differentiation Question
Hi,

I have been given the question differentiate the following equation logarithmically:

Attachment 20304

However, I don't yet understand factorising very well, and the answer given for this question has been factorised so I can't compare it with my own answer.

I took the following steps to work this out:

Attachment 20305

If someone can verify wherether this is correct or not, and tell me how if it's not, that would really useful.

Thanks.
• Jan 1st 2011, 12:51 PM
pickslides
Where did the 2 come from in your last line of working?

$\displaystyle \frac{e^x}{x}+e^x\ln x \neq \frac{2e^x}{x}+\ln x$
• Jan 1st 2011, 12:52 PM
It's all correct except for line 7, where $e^x$ is a factor.
• Jan 1st 2011, 12:56 PM
dwsmith
$\displaystyle y=\exp{(e^x\ln{x})}\Rightarrow \frac{dy}{dx}=\exp{(e^x\ln{x})}\left[e^x\ln{x}+e^x*\frac{1}{x}\right]$

$\displaystyle\Rightarrow x^{e^x}e^x\left[\ln{x}+\frac{1}{x}\right]$
• Jan 1st 2011, 01:24 PM
ashleysmithd
Ah I see, I tried factorising a little so the $\displaystyle \frac{e^x}{x}+e^x$ became $\displaystyle \frac{2e^x}{x}$

I realise this should be $\displaystyle \frac{2e^x * ln(x)}{x}$

and not

$\displaystyle \frac{2e^x}{x}$ $+ ln(x)$

When you say $e^x$ is a factor on line 7, which $e^x$ is this?
• Jan 1st 2011, 01:30 PM
Jameson
You see why that isn't true though right? $\displaystyle \frac{e^x}{x}+e^x \ne \frac{2e^x}{x}$ You can factor it however to $\displaystyle e^x \left( \frac{1}{x}+\ln(x) \right)$
• Jan 1st 2011, 01:46 PM
ashleysmithd
Ok I can see how that part is wrong now.

If I replaced step 7 with this:
http://djasmedia.com/rave/mathq3.jpg

Would it then be correct? I realise it can be factorised but, for now I'd rather work longhand and tackle factorising another day.
• Jan 1st 2011, 01:47 PM
dwsmith
Yes
• Jan 1st 2011, 01:54 PM
ashleysmithd
Brilliant thanks.