Question

If y=e^x^2 ln(x^2), find dy/dx

Answer 1

\[\huge y=e^{x^2}\cdot \ln(x^2)\]That's our y? :D Oh boy.

Answer 2

Sooooooo, product rule yes?
We have 2 separate thingies with x in them.\[\large \frac{dy}{dx}=\frac{d}{dx}\left[e^{x^2}\right]\ln(x^2)+e^{x^2}\frac{dy}{dx}\left[\ln(x^2)\right]\]Understand the setup? :)

Answer 3

not really

Answer 4

I prefer to use the PRIME notation, it's a little easier to read, let's see if that makes a difference.
Here is the definition of product rule,\[\huge (uv)'=u'v+uv'\]

Answer 5

\[\large \left(e^{x^2}\cdot \ln(x^2)\right)'=\left(e^{x^2}\right)'\cdot \ln(x^2)+e^{x^2}\cdot\left(\ln(x^2)\right)'\]
The primes are where we need to take a derivative.

Answer 6

It's a bit of a tricky problem! :O
If you can understand the setup then it's not too bad after that :D

Answer 7

so how do i take the derivative of e^x^2 and ln(x^2)

Answer 8

Remember that with an exponential of base e, the exponential won't change, but we'll apply the chain rule, multiplying by the derivative of the exponent that's up top.
\[\huge (e^{x^2})'=e^{x^2}(x^2)'\]
Giving us,\[\huge e^{x^2}(2x)\]Understand that one? :O

Answer 9

yes

Answer 10

Remember the derivative of natural log? :)\[\large (\ln (x)\;)'=\frac{1}{x}\]It's just important to remember that WHATEVER is inside of the log, is what you stuff into the denominator.\[\large (\ln(x^2)\;)'=\frac{1}{x^2}(x^2)'\]Then of course we have to apply the chain rule again :D\[\large =\frac{1}{x^2}(2x)\qquad =\qquad \frac{2}{x}\]