Answer:
[tex] \frac{d}{dx} (ln(e^{x^2}) ) = \frac{1}{e^{x^2}} 2x e^{x^2} = 2x[/tex]
Step-by-step explanation:
For this case we want to find the derivate of this function:
[tex] y = ln(e^{x^2}) )[/tex]
And in order to find the derivate we need to apply the chain rule given by:
[tex] \frac{df(u)}{dx} =\frac{df}{du} \frac{du}{dx}[/tex]
And on this case [tex] f = ln(u), u = e^{x^2}[/tex]
And we can find the partial derivates like this:
[tex] \frac{d}{du} (ln(u)) =\frac{1}{u} [/tex]
[tex] \frac{d}{dx}(e^{x^2})= e^{x^2} (2x) [/tex]
And if we replace we got:
[tex] \frac{d}{dx} (ln(e^{x^2}) ) = \frac{1}{u} 2x e^{x^2}[/tex]
And if we replace[tex]u = e^{x^2}[/tex] we got:
[tex] \frac{d}{dx} (ln(e^{x^2}) ) = \frac{1}{e^{x^2}} 2x e^{x^2} = 2x[/tex]
And that would be our final answer on this case
