1. ## Derivative Problem

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

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

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

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

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

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

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

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

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

Apparently you forgot the denominator...

Tonio

3. 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

4. It's technically never necessary to use the quotient rule because it's just a form of the product rule.

5. Ya i thought of that to after i did the problem.

6. $\frac{d}{dx} \ \frac{e^{x^{2}}}{1+ \ln(x)} = \frac{2xe^{x^{2}}}{1+\ln(x)} + e^{x^{2}} *\frac{-1}{(1+\ln(x))^{2}}*\frac{1}{x}$

7. Woah .... so wait.. our answers are the same correct you just simplified yours more correct?

8. 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.

9. 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.