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 jskrzy November 18th, 2012 12:34 PM

Differentiating e

I have been given the equation y=xe^3x

The answer is e^3x (1 + 3x)

How?

I thought maybe they are pulling down the 3x and and multiplying by 1, but why +1?

 MarkFL November 18th, 2012 12:53 PM

Re: Differentiating e

We are given to differentiate:

$y=xe^{3x}$

Using the product and chain rules, we find:

$\frac{dy}{dx}=x$$3e^{3x}$$+(1)e^{3x}=e^{3x}(3x+1)$

 jskrzy November 18th, 2012 12:58 PM

Re: Differentiating e

Oh I see. Thank you.

Another similar problem is e^(1+3t)^2

I just am not understanding how to get the 6(1+3t).

Thank you for the continued help.

 MarkFL November 18th, 2012 01:08 PM

Re: Differentiating e

Using the chain rule, we find:

$\frac{d}{dx}$$e^{(1+ 3t)^2}$$=e^{(1+3t)^2}\frac{d}{dx}$$(1+ 3t)^2$$=e^{(1+3t)^2}$$2(1+ 3t)\frac{d}{dx}(1+3t)$$=e^{(1+ 3t)^2}$$2(1+3t)(3)$$=$

$6(1+3t)e^{(1+ 3t)^2}$

 jskrzy November 18th, 2012 01:56 PM

Re: Differentiating e

Ah, I see that we must split it up many times.

Can you offer your help on this equation?

y=x^2 * lnx

and

t^2 + 2lnt

Thank you for helping. I am slowly understanding :)

 HallsofIvy November 19th, 2012 10:05 AM

Re: Differentiating e

Quote:
 Originally Posted by jskrzy Ah, I see that we must split it up many times. Can you offer your help on this equation? y=x^2 * lnx
What, exactly, do you want to do with this? Differentiate? Then use the product rule. $y'= (x^2)' ln(x)+ (x^2)(ln(x))#39;$.
Do you know the derivatives of $x^2$ and $ln(x)$?

and

t^2 + 2lnt

Thank you for helping. I am slowly understanding :)[/quote]

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