# Thread: Distributing the log function

1. ## Distributing the log function

Is distributing the natural log function legal?

EX: (d/dx) ln(x^2((cosx)^2)e^-x^2)

= (d/dx) ln(x^2) (d/dx) ln(cos(x)^2) (d/dx) -x^2

2. Not like that. You can use the log addition rule: $\ln(x^2 \cos^2(x) e^{-x^2}) = \ln(x^2) + \ln(\cos^2(x)) + \ln(e^{-x^2})$

All those can be simplified further using other log rules and can be differentiated separately but don't forget the chain rule when differentiating!"

3. Ah forgot about the log rules. Could someone check my work on this question.

d/dx ln(x^2((cos(x)^2)e^-x^2)

=(2x/x^2)+(1/((cos(x))^2))(-2cos(x)sin(x))+ e^-x^2

=((2x)-(2cos(x)sin(x))-2x(e^-x^2))/(x^2)((cos(x))^2)

Can't see a way I could simplify more, or even if the answer is right.

4. Originally Posted by ppark
Ah forgot about the log rules. Could someone check my work on this question.

d/dx ln(x^2((cos(x)^2)e^-x^2)

=(2x/x^2)+(1/((cos(x))^2))(-2cos(x)sin(x))+ e^-x^2
ln(e^{-x^2})= -x^2. Its derivative is -2x.

=((2x)-(2cos(x)sin(x))-2x(e^-x^2))/(x^2)((cos(x))^2)
I am not sure how you have combined these. $\frac{a}{b}+ \frac{c}{d}$ is NOT equal to $\frac{a+ c}{bd}$.

Can't see a way I could simplify more, or even if the answer is right.