# Thread: Logarithmic Differentiation Question

1. ## Logarithmic Differentiation Question

Hi,

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

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:

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

Thanks.

2. Where did the 2 come from in your last line of working?

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

3. It's all correct except for line 7, where $\displaystyle e^x$ is a factor.

4. $\displaystyle \displaystyle y=\exp{(e^x\ln{x})}\Rightarrow \frac{dy}{dx}=\exp{(e^x\ln{x})}\left[e^x\ln{x}+e^x*\frac{1}{x}\right]$

$\displaystyle \displaystyle\Rightarrow x^{e^x}e^x\left[\ln{x}+\frac{1}{x}\right]$

5. Ah I see, I tried factorising a little so the $\displaystyle \displaystyle \frac{e^x}{x}+e^x$ became $\displaystyle \displaystyle \frac{2e^x}{x}$

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

and not

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

When you say $\displaystyle e^x$ is a factor on line 7, which $\displaystyle e^x$ is this?

6. You see why that isn't true though right? $\displaystyle \displaystyle \frac{e^x}{x}+e^x \ne \frac{2e^x}{x}$ You can factor it however to $\displaystyle \displaystyle e^x \left( \frac{1}{x}+\ln(x) \right)$

7. Ok I can see how that part is wrong now.

If I replaced step 7 with this:

Would it then be correct? I realise it can be factorised but, for now I'd rather work longhand and tackle factorising another day.

8. Yes

9. Brilliant thanks.