Finding Derivatives with Natural Logarithms

[Swerting]

Hello, I was given a problem dealing with derivatives and compositions of g and f. The problem is, when I go to find the derivative at a certain point, it involves natural logs. I can only get so far when I can no longer get 'h' out of the equation!
f(x)=ln(x) for all x>0
g(x)=(x^2)-4
H(x)=f(g(x))=2ln(x-2)
The equation for a derivative is (f(x+h)-f(x))/(h) as h approches 0.
When I attempted to solve H'(7), this is what I got.
$$H'(7)=\frac{H(7+h)-H(7)}{h}\ \mbox{as h approaches 0}$$
$$H'(7)=\frac{2ln((7+h)-2)-2ln(7-2)}{h}\ \mbox{as h approaches 0}$$
$$H'(7)=\frac{2ln(5+h)-2ln(5)}{h}\ \mbox{as h approaches 0}$$
$$H'(7)=\frac{2(ln(5+h)-ln(5))}{h}\ \mbox{as h approaches 0}$$

Again, I am trying to get the 'h' out of the natural log, meaning next to the '2' so I can cancel out the h, but am having trouble.
I know that there is a 'shortcut' and that the answer should be $$\frac{2}{5}$$, and it is probably a simple manipulation of the natural logs, but I've asked around and no one can seem to help me. I thank you for your help.

-Swerting

 

[HallsofIvy]

Hello, I was given a problem dealing with derivatives and compositions of g and f. The problem is, when I go to find the derivative at a certain point, it involves natural logs. I can only get so far when I can no longer get 'h' out of the equation!
f(x)=ln(x) for all x>0
g(x)=x^2
H(x)=f(g(x))=2ln(x-2)

No. H(x)= ln(x²)= 2 ln(x)

The equation for a derivative is (f(x+h)-f(x))/(h) as h approches infinity.

No, it isn't. It is the limit as h approaches 0.

When I attempted to solve H'(7), this is what I got.
$$H'(7)=\frac{H(7+h)-2-H(7)}{h}\ \mbox{as h approaches 0}$$

No. (H(7+h)- H(7))/h = (ln((7+h)²- ln(49))/h= 2(ln(7+h)- ln(7))/h

$$H'(7)=\frac{2ln((7+h)-2)-2ln(7-2)}{h}\ \mbox{as h approaches 0}$$
$$H'(7)=\frac{2ln(5+h)-2ln(5)}{h}\ \mbox{as h approaches 0}$$
$$H'(7)=\frac{2(ln(5+h)-ln(5))}{h}\ \mbox{as h approaches 0}$$

Again, I am trying to get the 'h' out of the natural log, meaning next to the '2' so I can cancel out the h, but am having trouble.
I know that there is a 'shortcut' and that the answer should be $$\frac{2}{5}$$,

And, there, together will all of the mistakes I pointed out before is your problem. The derivative of ln(x²) at x= 7 is 2/7, not 2/5.

and it is probably a simple manipulation of the natural logs, but I've asked around and no one can seem to help me. I thank you for your help.

-Swerting

You can't get the "h" out of the natural log. Do you know the standard proof of the derivative of ln(x) itself?

Are you required to use that definition? I would think that using the derivative of ln(x) together with the chain rule of using the fact that if y= 2 ln(x) then x= e^y/2 would be much simpler.

 

[Swerting]

Pardon me, the text on the page is quite small, and I overlooked something quite important! g(x)=(x^2)-4, that is my fault, but that is how I reached H(x)=2ln(x-2). Also, the '-2' in the first line of my attempt shouldn't be there, but I can't seem to get rid of it in the post. Finally, yes, I assume I am required to use that definition. We are not that far into the material, and we only know the one equation, and what H(x) equals.

P.S.- Sorry, the first post is always overwhelming to me, maybe double checking isn't enough anymore, next time I will triple check.