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 November 18th, 2012, 11:34 AM #1 Member   Joined: Apr 2011 Posts: 47 Thanks: 0 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?
 November 18th, 2012, 11:53 AM #2 Senior Member     Joined: Jul 2010 From: St. Augustine, FL., U.S.A.'s oldest city Posts: 12,204 Thanks: 511 Math Focus: Calculus/ODEs 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)$
 November 18th, 2012, 11:58 AM #3 Member   Joined: Apr 2011 Posts: 47 Thanks: 0 Re: Differentiating e Oh I see. Thank you. Another similar problem is e^(1+3t)^2 The answer is 6(1+3t)*e^(1+3t)^2 I just am not understanding how to get the 6(1+3t). Thank you for the continued help.
 November 18th, 2012, 12:08 PM #4 Senior Member     Joined: Jul 2010 From: St. Augustine, FL., U.S.A.'s oldest city Posts: 12,204 Thanks: 511 Math Focus: Calculus/ODEs 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}$$2(1+3t)(3)$$=$ $6(1+3t)e^{(1+ 3t)^2}$
 November 18th, 2012, 12:56 PM #5 Member   Joined: Apr 2011 Posts: 47 Thanks: 0 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
November 19th, 2012, 09:05 AM   #6
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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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