I am just learning how to do these problems. I have to find the derivative of y with respect to t for
y= ln (t^2)
Can someone explain the formula for doing a problem like this?
You are supposed to know that the derivative of $\displaystyle f(x)=\ln(x)$ is $\displaystyle f'(x)=\frac1x$.
Since you don't have a single variable t you have to use
a) some laws of logaritm: $\displaystyle y = \ln(t^2)~\implies~y=2\ln(t)~\implies~ y'=\frac2t$
b) the chain rule: $\displaystyle y = \ln(t^2)~\implies~y'=\frac1{t^2} \cdot 2t ~\implies~ y'=\frac2t$
Be aware that the logrithm function is only defined for positive arguments.
Hi KarlosK, so taking the derivative of a natural logarithmic function is simply $\displaystyle \frac{1}{x} $. In your case your $\displaystyle x $ would be simply $\displaystyle t^2 $. By chain rule, after taking the derivative of the first function, theres is a function within a function which is the $\displaystyle t^2 $, so there for it becomes $\displaystyle 2t $. Yet, yielding the final solution as $\displaystyle y' = \frac{2}{t} $.