Finding the first derivative of a equation.

[Amel]

Homework Statement

Estimate f ' (2) for 7^x

Homework Equations

[f(x+h)-f(x)]/h

The Attempt at a Solution

Ok so I attempted to do this. I put in [(7^(2+h))-7²]/h
I must be messing up with my algebra. SO then take log of both
I got
(2+h)log7-log49 all over h
distributed log7 to (2+h) and got 1.69+ .845h - 1.69 all over h
the 1.69 canceled and I have (.845h)/h left and that can't be right because then I cancel h and got no h left to plug in for zero.

What do I do obviously I am doing it wrong.

 

[Mark44]

Homework Statement

Estimate f ' (2) for 7^x

Homework Equations

[f(x+h)-f(x)]/h

The Attempt at a Solution

Ok so I attempted to do this. I put in [(7^(2+h))-7²]/h
I must be messing up with my algebra. SO then take log of both
I got

Taking the log is incorrect. There is no property of logs that says log(a + b) = log a + log b.

Instead, what you want to do is to factor 7² from the two terms in the numerator, giving you
\frac{7^2(7^h - 1)}{h}
To find f'(2) you need to evaluate this limit:
\lim_{h \rightarrow 0}\frac{7^2(7^h - 1)}{h}~=~49\lim_{h \rightarrow 0}\frac{7^h - 1}{h}
If you know L'Hopital's Rule, this limit is not too hard to evaluate, but if you don't know it, it's fairly difficult to evaluate. The expression we're taking the limit of is indeterminate of the form [0/0]. It's called indeterminate because you can't determine what the limit will actually be without some work, and can come out to be anything.

Before continuing, I need to know whether you are able to find this limit for yourself.

(2+h)log7-log49 all over h
distributed log7 to (2+h) and got 1.69+ .845h - 1.69 all over h
the 1.69 canceled and I have (.845h)/h left and that can't be right because then I cancel h and got no h left to plug in for zero.

What do I do obviously I am doing it wrong.