Finding a derivative through the limit method

[lyarbrough]

Homework Statement

Hi, I'm trying to understand how you solve for the problem lim [(2+h)^5 -2^5]/h as h→0
I already have the solution, but I want to make sure my understanding of how it's done is correct.

Homework Equations

I'm suppose to be using the definition of the derivative [f(x+h)-f(x)]/h

The Attempt at a Solution

So what I have is that lim [(2+h)^5 -2^5]/h as h→0 is = f'(2). I'm assuming that's because 2 was substituted in for f(x) in the definition of a derivative.

The next step I have is that f(x)= ? and then f'(x) = x^5. I'm just wondering if it's just re substituting in x for the 2? I know how to solve it from there with the chain rule, I'm just wondering how they determine f'(x). Thanks!

 

[Dick]

Homework Statement

Hi, I'm trying to understand how you solve for the problem lim [(2+h)^5 -2^5]/h as h→0
I already have the solution, but I want to make sure my understanding of how it's done is correct.

Homework Equations

I'm suppose to be using the definition of the derivative [f(x+h)-f(x)]/h

The Attempt at a Solution

So what I have is that lim [(2+h)^5 -2^5]/h as h→0 is = f'(2). I'm assuming that's because 2 was substituted in for f(x) in the definition of a derivative.

The next step I have is that f(x)= ? and then f'(x) = x^5. I'm just wondering if it's just re substituting in x for the 2? I know how to solve it from there with the chain rule, I'm just wondering how they determine f'(x). Thanks!

Since you say you know the answer, the limit of that is 80. That's the answer. What's the question? Are they asking you to guess what f(x) would give you that difference quotient to determine f'(2)? That shouldn't be hard. Are you sure they didn't say f(x)=x^5, not f'(x)=x^5?

 

[Chestermiller]

Hi lyrabrough. Welcome to Physics Forums.

The way they want you to do this problem is not by actually taking the derivative. They want you to find the limit of that ratio as h approaches zero.

What you need to do is factor that numerator into two factors. One of the factors will cancel with the h in the denominator. The other factor will be evaluated at h = 0. Consider this:

x⁵-y⁵=(x-y)(x⁴+x³y+x²y²+xy³+y⁴)

Chet