Why the Derivative of e^x Is e^x: Proof of the Exponential Derivative
The function f(x) = e^x is a calculus student's favorite because it's equal to its own derivative, and as a consequence of that fact, it's also equal to its own antiderivative. But you may wonder why d/dx[e^x] = f'(x) = e^x, and why it is the only function that equals its own derivative.
There are two ways to prove that [e^x]' = e^x. One is to apply the limit definition of the derivative to e^x, and then simplify this expression by applying the limit definition of e^x. The second method of proof is to take power series definition of e^x and differentiate it term by term. With both methods you end up with the same function you started with, thus proving that e^x is its own derivative.
If you plug the function f(x) = e^x into the derivative limit equation, you get
= lim(h→0) [e^(x+h) - e^x]/h
= lim(h→0) [(e^x)(e^h - 1)]/h
= (e^x) * lim(h→0) [e^h - 1]/h
The line immediately above shows that the derivative of e^x is e^x times a constant that is equal to lim(h→0) [e^h - 1]/h. But what is the value of lim(h→0) [e^h - 1]/h? We cannot plug h directly into this expression because it will yield the indeterminate form 0/0. Instead, we must first break down e^h using the limit definition of the exponential function.
The limit definition of the e^h is
e^h = lim(N→∞) [1 + h/N]^N
= lim(N→∞) [ 1 + B(N, 1)(h/N) + B(N, 2)(h/N)^2 + ... + B(N, N)(h/N)^N ]
where B(N, k) is the binomial coefficient given by
B(N, k) = N!/[k!(N-k)!]
The value of B(N, 1) is N. If we plug the expanded version of e^h into the expression (e^h - 1)/h, we get
(e^h - 1)/h
= lim(N→∞) [ 1 + N(h/N) + B(N, 2)(h/N)^2 + ... + B(N, N)(h/N)^N - 1]/h
= lim(N→∞) [N(h/N) + B(N, 2)(h/N)^2 + ... + B(N, N)(h/N)^N]/h
= lim(N→∞) [h + B(N, 2)(h/N)^2 + ... + B(N, N)(h/N)^N]/h
= lim(N→∞) 1 + [B(N, 2)/N^2]*h + [B(N, 3)/N^3]*h^2 + ... + [B(N, N)/N^N]*h^(N-1)
The limit of the coefficients [B(N, k)/N^k] is finite as N goes to infinity. Therefore, if we plug h = 0 directly into the expression above, we get the sum
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1 + 0 + 0 + 0 + ... + 0 = 1
Thus, we have
lim(h→0) [e^h - 1]/h = 1.
Now if we take this result to the first several lines of this prove we see that
= (e^x) * lim(h→0) [e^h - 1]/h
= (e^x) * 1
And the proof is complete.
The power series definition of the exponential function e^x is
e^x = 1 + x + (x^2)/2 + (x^3)/6 + (x^4)/24 + ... + (x^n)/n! + ...
The derivative of this infinite sum is the infinite sum of the derivatives of the individual terms. This gives us
= 0 + 1 + 2(x^1)/2 + 3(x^2)/6 + 4(x^3)/24 + ... + n(x^(n-1))/n! + ...
= 1 + x + (x^2)/2 + (x^3)/6 + ... + (x^(n-1))/(n-1)! + ...
The power series in the line immediately above is exactly the power series of e^x. Thus the proof is complete and we can see that e^x is its own derivative. This proof is much shorter than the previous proof!
Bonus: Proof 3
There is yet a third way to prove the derivative of e^x using the derivative of Ln(x) (proved in another tutorial) and the chain rule. We start with the facts that
- Ln(e^x) = x
- d/dx[ Ln(x) ] = 1/x
- d/dx[ x ] = 1
Therefore we have
1 = d/dx[ x ]
= d/dx[ Ln(e^x) ]
= (1/e^x) * d/dx[ e^x ], by the chain rule
Solving the expression 1 = (1/e^x) * d/dx[ e^x ] gives us e^x = d/dx[ e^x ].
Example of How to Use the Derivative of e^x
Let's find the equation of the tangent line on the graph of y = e^x at the point x = 0.75. We know that one point on this line is (0.75, e^0.75) ≈ (0.75, 2.117), because this is where the line touches the graph of the exponential curve.
We also know that the slope of the tangent line is e^0.75 ≈ 2.117. Therefore, the equation of the tangent line is
y - y0 = m(x - x0)
y - 2.117 = 2.117(x - 0.75)
y = 2.117x + 0.52925