# Thread: Derivatives of natural logarithms

1. ## Derivatives of natural logarithms

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?

2. Originally Posted by KarlosK
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 $f(x)=\ln(x)$ is $f'(x)=\frac1x$.

Since you don't have a single variable t you have to use

a) some laws of logaritm: $y = \ln(t^2)~\implies~y=2\ln(t)~\implies~ y'=\frac2t$

b) the chain rule: $y = \ln(t^2)~\implies~y'=\frac1{t^2} \cdot 2t ~\implies~ y'=\frac2t$

Be aware that the logrithm function is only defined for positive arguments.

3. Originally Posted by KarlosK
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?
Dear KarlosK,

You have to use,

1) $\frac{d}{dx}lnx=\frac{1}{x}$

2)The Chain rule

4. Originally Posted by earboth
You are supposed to know that the derivative of $f(x)=\ln(x)$ is $f'(x)=\frac1x$.

Since you don't have a single variable t you have to use

a) some laws of logaritm: $y = \ln(t^2)~\implies~y=2\ln(t)~\implies~ y'=\frac2t$

b) the chain rule: $y = \ln(t^2)~\implies~y'=\frac1{t^2} \cdot 2t ~\implies~ y'=\frac2t$

Be aware that the logrithm function is only defined for positive arguments.
In part a, I am having trouble seeing how you went from y=2 ln(t) to the y'= 2/t can you just explain that one step.

Also do you need to do both parts a/b, or is that just a way to check the answer? Thanks for the help!!

5. Originally Posted by KarlosK
In part a, I am having trouble seeing how you went from y=2 ln(t) to the y'= 2/t can you just explain that one step.

Also do you need to do both parts a/b, or is that just a way to check the answer? Thanks for the help!!
Dear KarlosK,

Do you know that, $\frac{d}{dx}lnx=\frac{1}{x}$ Thats what he had used when going from y=2 ln(t) to the y'= 2/t.

a) and b) are two methods of doing the same problem. You can do it anyway you wish.

Hope this helps.

6. Originally Posted by KarlosK
In part a, I am having trouble seeing how you went from y=2 ln(t) to the y'= 2/t can you just explain that one step.

Also do you need to do both parts a/b, or is that just a way to check the answer? Thanks for the help!!
Hi KarlosK, so taking the derivative of a natural logarithmic function is simply $\frac{1}{x}$. In your case your $x$ would be simply $t^2$. By chain rule, after taking the derivative of the first function, theres is a function within a function which is the $t^2$, so there for it becomes $2t$. Yet, yielding the final solution as $y' = \frac{2}{t}$.