Question

Please help.
Use the definition of the derivative to express g'(x) as a limit if g(x)=(lnx)^(2x). Thanks in advance.

Answer 1

\[\lim_{H \rightarrow 0} \frac{ F(X+H)-F(X) }{ H }\]

Answer 2

If all you need to do is express the derivative as a limit, then you need only replace \(f(x+h)\) and \(f(x)\) in the definition of the derivative and you're done.
\[g'(x)=\lim_{h\to0}\frac{g(x+h)-g(x)}{h}=\cdots\]
If \(g(x)=(\ln x)^{2x}\), what is \(g(x+h)\) ?

Answer 3

(lnx+h)^(2x+h)

Answer 4

and you would subtract the original g(x) from that, all divided by H?

Answer 5

Close. You're not simply replacing each \(x\) with \(x\) then adding \(h\); you are replacing each \(x\) with the expression \(x+h\), like so:
\[g(x)=(\ln x)^{2x}~~\implies~~g(\color{red}{x+h})=(\ln(\color{red}{x+h}))^{2(\color{red}{x+h})}=(\ln(x+h))^{2x+2h}\]

Answer 6

ok gotcha! now you would subtract the original g9x) from that, correct? and its all divided by h.

Answer 7

i meant g(x)

Answer 8

Correct, and don't forget that this is a limit we're dealing with here. \(\lim\limits_{h\to0}\cdots\)

Answer 9

\[\frac{ (\ln(x+h))^(2x+2h)-(lnx)^(2x) }{ h }\]

Answer 10

yep, so it would look like this. those are supposed to be the exponents, not sure why they weren't raised.

Answer 11

now where do we go from here?

Answer 12

When you want multiple-character exponents (i.e. expressions, numbers with more than one digit, etc) you have to group them properly in latex with curly braces. Compare `1^2`, `1^23`, and `1^{23}`: \(1^2\), \(1^23\), and \(1^{23}\).

Anyway, like I said, the question simply asks for a limit form of the derivative of \(g(x)\), so you're done once you have
\[\lim_{h\to0}\frac{(\ln(x+h))^{2x+2h}-(\ln x)^{2x}}{h}\]
Incidentally, there is a simpler form, but I don't know if you're expected to do that.

Answer 13

ok that looks better l9ol. Thank you very much I appreciate it!!