Simplifying equation in to stages to follow...

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June 24th 2012, 06:18 AM
Cyclo

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 ?

Thanks for your help.
June 24th 2012, 07:50 AM
earboth

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

Quote:

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 ?

Thanks for your help.

Please use parantheses for grouping!

Which one of the following equations do you mean:

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

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

y= e^(x^(2* ln (x))) ---- > $\displaystyle{y = e^{x^{(2 \cdot \ln(x))}}$
June 24th 2012, 09:11 AM
Cyclo

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

Quote:

Originally Posted by earboth

Please use parantheses for grouping!

Which one of the following equations do you mean:

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

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

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

Apologies. I mean the first one.
June 24th 2012, 09:38 AM
Reckoner

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

Quote:

Originally Posted by Cyclo

Apologies. I mean the first one.

$y = e^{x^2}\cdot\ln x$

Use the product rule:

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

Can you finish?
June 24th 2012, 10:14 AM
Cyclo

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

Quote:

Originally Posted by Reckoner

$y = e^{x^2}\cdot\ln x$

Use the product rule:

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

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
June 24th 2012, 10:20 AM
Reckoner

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

Quote:

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:

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

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