Hi Guys,

I'm a little stuck on how on to differentiate when equations involve e and ln. How would I go about differentiating y= e^x^2 ln x for instance ?

Thanks for your help.

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- Jun 24th 2012, 06:18 AMCycloSimplifying equation in to stages to follow...
Hi Guys,

I'm a little stuck on how on to differentiate when equations involve e and ln. How would I go about differentiating y= e^x^2 ln x for instance ?

Thanks for your help. - Jun 24th 2012, 07:50 AMearbothRe: Simplifying equation in to stages to follow...
Please use parantheses for grouping!

Which one of the following equations do you mean:

y= e^(x^2)*ln (x) ---- > $\displaystyle \displaystyle{y = e^{x^2} \cdot \ln(x)}$

y= e^((x^2)* ln (x)) ---- > $\displaystyle \displaystyle{y = e^{x^2 \cdot \ln(x)}}$

y= e^(x^(2* ln (x))) ---- > $\displaystyle \displaystyle{y = e^{x^{(2 \cdot \ln(x))}}$ - Jun 24th 2012, 09:11 AMCycloRe: Simplifying equation in to stages to follow...
- Jun 24th 2012, 09:38 AMReckonerRe: Simplifying equation in to stages to follow...
- Jun 24th 2012, 10:14 AMCycloRe: Simplifying equation in to stages to follow...
- Jun 24th 2012, 10:20 AMReckonerRe: Simplifying equation in to stages to follow...
You can't just take the derivative of one factor and multiply it by the derivative of the other factor. As I said, we have to use this product rule:

$\displaystyle \frac d{dx}[uv] = u\frac{dv}{dx} + v\frac{du}{dx}.$

So your answer, before simplification, should have two terms.