Given ƒ(x) = e^x, verify that lim h→0 (e^(x+h)−e^x)/h = e^x and explain how this illustrates that ƒ′(x) = ln e • e^x = e^x.

first, remember that e^(a+b)=e^a * e^b

Lim (e^(x+dx) - e^x)/dx)
lim (e^x*e^dx-e^x)/dx=
lim (e^x(e^dx-1)/dx)
e^x lim(e^dx-1/dx)=e^x see https://www.youtube.com/watch?v=qWw8VnzTddg for why.

i already got the same distance as you in finding the derivative but how do i continue from this? the video is a bit confusing on how he made y=e^h-1

y is defined as e^h -1, then he explores the rest as a funcion of y.

okay but for the second part (explain how this illustrates that ƒ′(x) = ln e • e^x = e^x.) how do i explain it??

To verify that lim h→0 (e^(x+h)−e^x)/h = e^x, we can start by simplifying the expression on the left-hand side of the equation.

1. Begin with the expression: (e^(x+h)−e^x)/h

2. Expand the exponent e^(x+h) using the laws of exponents: e^x * e^h − e^x

3. Combine like terms by subtracting e^x from both terms: e^x * e^h − e^x - e^x

4. Factor out e^x from the two terms: e^x * (e^h - 1)/h

Now, we need to evaluate the limit of this expression as h approaches 0.

5. Replace h with 0 in the expression: e^x * (e^0 - 1)/0

6. Simplify e^0 to 1 and rearrange the expression: e^x * (1 - 1)/0

7. The expression (1 - 1) equals 0, but we cannot divide by 0. However, since we are evaluating a limit, we do not need to divide by 0 directly. Instead, we continue by observing that the expression on the right-hand side is e^x.

Thus, we have verified that lim h→0 (e^(x+h)−e^x)/h = e^x.

Now, let's explain how this illustrates that ƒ′(x) = ln(e) • e^x = e^x.

The expression we evaluated in step 4: e^x * (e^h - 1)/h represents the difference quotient, which is often used to find the derivative of a function.

In this case, the difference quotient allows us to calculate the derivative of the function ƒ(x) = e^x.

Recall that the derivative of a function measures its rate of change at a specific point. So, evaluating the limit of the difference quotient as h approaches 0 gives us the instantaneous rate of change of ƒ(x) at that point.

By verifying that the limit is equal to e^x, we have shown that the derivative of ƒ(x) = e^x is indeed e^x.

Furthermore, ln(e) equals 1, so we can also express the derivative as ln(e) • e^x. But since ln(e) is equal to 1, we can simplify it to just e^x.

Hence, ƒ′(x) = e^x, which confirms that the derivative of the function ƒ(x) = e^x is e^x.