# Derivative Problem

• Feb 21st 2010, 11:33 AM
x5pyd3rx
Derivative Problem
Im pretty sure i did the derivative part right but i just want a second take on my algerbra

$\displaystyle \frac {e^{x^2}}{1+ \ln x}$

$\displaystyle \frac {dy}{dx} = 1 + \ln x * e^{x^2} *2x - e^{x^2} * \frac {1}{x}$

$\displaystyle = 2x(1 + \ln x ) * e^{x^2} - \frac {e^{x^2}}{x}$
• Feb 21st 2010, 11:36 AM
tonio
Quote:

Originally Posted by x5pyd3rx
Im pretty sure i did the derivative part right but i just want a second take on my algerbra

$\displaystyle \frac {e^{x^2}}{1+ \ln x}$

$\displaystyle \frac {dy}{dx} = 1 + \ln x * e^{x^2} *2x - e^{x^2} * \frac {1}{x}$

$\displaystyle = 2x(1 + \ln x ) * e^{x^2} - \frac {e^{x^2}}{x}$

It's wrong: the quotient derivative's rule is $\displaystyle \left(\frac{u}{v}\right)'=\frac{u'v-uv'}{v^2}$.

Apparently you forgot the denominator...(Worried)

Tonio
• Feb 21st 2010, 11:40 AM
x5pyd3rx
Jeez i feel dumb but the algerbra on top is correct and all i would do is throw the (1 + ln x)^2 on the bottom.

you would think someone taking calc 2 would know the quotient rule (Speechless)
• Feb 21st 2010, 11:52 AM
Random Variable
It's technically never necessary to use the quotient rule because it's just a form of the product rule.
• Feb 21st 2010, 11:55 AM
x5pyd3rx
Ya i thought of that to after i did the problem.
• Feb 21st 2010, 12:05 PM
Random Variable
$\displaystyle \frac{d}{dx} \ \frac{e^{x^{2}}}{1+ \ln(x)} = \frac{2xe^{x^{2}}}{1+\ln(x)} + e^{x^{2}} *\frac{-1}{(1+\ln(x))^{2}}*\frac{1}{x}$
• Feb 21st 2010, 12:16 PM
x5pyd3rx
Woah .... so wait.. our answers are the same correct you just simplified yours more correct?
• Feb 21st 2010, 12:24 PM
Random Variable
Quote:

Originally Posted by x5pyd3rx
Woah .... so wait.. our answers are the same correct you just simplified yours more correct?

Yes, our answers are the same. I can simplify mine so that it looks like yours. I just wanted to show how you can use the product rule instead of the quotient rule.
• Feb 21st 2010, 12:36 PM
vince
Quote:

Originally Posted by Random Variable
Yes, our answers are the same. I can simplify mine so that it looks like yours. I just wanted to show how you can use the product rule instead of the quotient rule.

back in school i always used the product rule to get around the quotient rule. but as of now, i am indifferent to them and consider the quotient rule to be so damn simple to remember.