Derivative of an exponential 2

In this problem, we’ll tackle the derivative of f(x)=7^x using the definition of the derivative.

1. Write down the difference quotient for f.
2. Rewrite your difference quotient from part 1 as a product of two terms: 1 involving only x and the other involving only h.
3. Using your result from part 2 and the table below, write down an approximate formula for f'(x).

1.f(x)=\frac{7^{x+h}-7^x}{h}
2.f(x)=\frac{7^x(7^h-1)}{h}
-f(x)=\frac{7^x}{h}*\frac{7^h-1}{h}

2.revised
f(x)=\frac{7^{x+h}-7^x}{h}
f(x)=\frac{7^x(7^h-1)}{h}
f(x)=7^x*\frac{7^h-1}{h}

3.f'(x)=7^x(1.94591)

I am confused on how to remove the h from the denominator of \frac{7^x}{h}. I know that the answer should be 7^x(ln(7)) and ln(7)\approx1.9459101. I cant figure out how to get it into 7^x*\frac{7^h-1}{h} format.

Well, I don’t think there should be an h in the denominator there. Note that

It’s not the case (as you seem to have used) that

I am a little confused on part 3. I know the rule \frac{d}{dx}a^x=a^x\times \ln (a) and how we ended part 2 using the rule \frac{d}{dx}a^x=a^x\times C_a{} but I am lost with how the \frac{\left(7^h-1\right)}{h} turns into ln(7)?

Is that just a known step or did you somehow use the table to get that value by substituting in a given h value?

@jbarry1 The general exponential rule that you refer to, namely \frac{d}{dx}a^x=a^x\times \ln (a), is still a little bit in our future and not immediately relevant to the problem at hand. That rule provides an exact formula for the derivative of a a^x but we don’t really have the tools to explain why it should work just yet. This problem asks us to derive an estimate to the derivative of a^x using tools that we have a firm grip on - namely the definition of the derivative and the numerical approximation of limits. So I guess that @mearing referred to the table given in the problem to surmise that

Now, it just so happens that \ln(7) \approx 1.94591, but maybe that’s just a coincidence.