# Thread: Simplifying equation in to stages to follow...

1. ## Simplifying 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 ?

2. ## Re: Simplifying equation in to stages to follow...

Originally Posted by Cyclo
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 ?

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

3. ## Re: Simplifying equation in to stages to follow...

Originally Posted by earboth

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))) ---- > $\displaystyle \displaystyle{y = e^{x^{(2 \cdot \ln(x))}}$
Apologies. I mean the first one.

4. ## Re: Simplifying equation in to stages to follow...

Originally Posted by Cyclo
Apologies. I mean the first one.
$\displaystyle y = e^{x^2}\cdot\ln x$

Use the product rule:

$\displaystyle \frac{dy}{dx} = e^{x^2}\cdot\frac d{dx}\left[\ln x\right] + \ln x\cdot\frac d{dx}\left[e^{x^2}\right]$

Can you finish?

5. ## Re: Simplifying equation in to stages to follow...

Originally Posted by Reckoner
$\displaystyle y = e^{x^2}\cdot\ln x$

Use the product rule:

$\displaystyle \frac{dy}{dx} = e^{x^2}\cdot\frac d{dx}\left[\ln x\right] + \ln x\cdot\frac d{dx}\left[e^{x^2}\right]$

Can you finish?
I'm not sure I have the full understanding of the terms, but here goes...

I'll have to read up on how to use the correct math text.

2x*e^xsquared*1/x

6. ## Re: Simplifying equation in to stages to follow...

Originally Posted by Cyclo
I'm not sure I have the full understanding of the terms, but here goes...

I'll have to read up on how to use the correct math text.

2x*e^xsquared*1/x
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:

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