Derivative of ln(x^(-2))

[MarkSA]

Hello,

Find the derivative of:
ln(x^(-2))

This is a practice problem for our physics math minitest in a few days.. but we've never learned how to find the derivative of this in calculus yet. What is the general procedure in solving derivatives of ln's or log's? I think it was mentioned in class the answer is -2/x but i'm not sure how that's arrived at.

 

[o_O]

Well first, you could use the rule:

\(\displaystyle log_{a} b^{n} = nlog_{a}b\)

And take the derivative accordingly.

Or you could use the chain rule (if you learned it) which states:

\(\displaystyle [f(g(x))]' = f'(g(x)) \cdot g'(x)\)

Where f(x) = ln(x) and g(x) = x[sup:3pg7xxjf]-2[/sup:3pg7xxjf]

 

[MarkSA]

Sorry, I meant that I don't know how to take the derivative of any log/ln. I'm familiar with the chain rule, but I don't know what to do when taking the derivative of log's/ln's.

We just finished the section on derivatives of e^. ie: if f(x) = e^(-3x), then f'(x) = -3e^(-3x)

But we've never done any derivatives with log's or ln's yet. Is this process similar for those?

How would one find the derivative of ln(x^(-2)) and something like: log base 2 of: x^3?

 

[o_O]

Ooh so you haven't gone over them yet. Well, without going through the proofs, the derivatives of lnx and more generally log[sub:2ybkokie]a[/sub:2ybkokie]x are:

\(\displaystyle \frac{d}{dx} lnx = \frac{1}{x}\)

\(\displaystyle \frac{d}{dx} log_{a}x = \frac{1}{xlna}\)

 

[MarkSA]

Thanks. So for the initial example, we would get this?

d/dx (ln(x^(-2)) = (1/x) * -2x^(-3)
= -2/x^4

Was that done correctly? It's a different answer than when it was worked out in class but I may have misread it (I was a ways off from the board)

edit: Or wait.. would the proper way to do it be:
= (1/x^(-2)) * -2x^(-3)
= -2/x. That's it I bet?

 

[o_O]

Yes, the latter is correct. Just the use of the chain rule.