Why does d/dx[Ln(x)] = 1/x? Proof That the Derivative of Ln(x) Equals 1/x
The derivative of the natural logarithm function Ln(x) is one of the first derivatives you learn in Calculus I. Unfortunately, not everyone who takes calculus learns why the derivative of Ln(x) is 1/x. If you took a college course like applied calculus for non-majors, you might have been required to memorize a list of derivatives to compute the slopes of tangent lines, but never learned where the derivative expressions came from. But as long as you learned the limit definition of the derivative
f'(x) = lim(h→0) [f(x+h) - f(x)]/h
and the definition of the mathematical constant e, which is the base of the natural logarithm
e = lim(n→∞) (1 + 1/n)^n ≈ 2.718281828459, and
e^z = lim(n→∞) (1 + z/n)^n
then you can understand the proof of the derivative of the natural logarithm function.
The first step in proving that the derivative of Ln(x) is 1/x is to apply the limit definition of the derivative to the function f(x) = Ln(x) and then use the fundamental properties of logarithms to simplify the expression. This gives us
= lim(h→0) [ Ln(x+h) - Ln(x) ] / h
= lim(h→0) [ Ln((x+h)/x) ] / h
= lim(h→0) [ Ln(1 + h/x) ] / h
= lim(h→0) (1/h) * Ln(1 + h/x)
= lim(h→0) Ln[ (1 + h/x)^(1/h) ]
= lim(h→0) Ln[ (1 + [1/x]/[1/h])^(1/h) ]
One of the nice properties of limits is that you can put them inside functions. In other words, if you have lim(x→a) f(x) you can rewrite it as f(lim(x→a)). This gives us equality
lim(h→0) Ln[ (1 + [1/x]/[1/h])^(1/h) ] = Ln[ lim(h→0) (1 + [1/x]/[1/h])^(1/h) ].
The next step is to make a change of variable, replacing h with 1/N, which implies that 1/h = N. As h goes to 0, N goes to infinity. Thus we can rewrite our limit equation in terms of N rather than h:
Ln[ lim(h→0) (1 + [1/x]/[1/h])^(1/h) ] ≡ Ln[ lim(N→∞) (1 + [1/x]/N)^N ]
Luckily for us, the limit expression on the right-hand side of the equivalence is recognizable as the definition of the function e^(1/x). This lets us simplify the expression even more from its original form. We now have
Ln [ lim(N→∞) (1 + [1/x]/N)^N ] = Ln[ e^(1/x) ]
Since the natural logarithm function and the exponential function are inverses of each other, they cancel out. Thus, the whole thing collapses to
Ln[ e^(1/x) ] = 1/x
and the proof of the derivative of the natural logarithm is complete.
Example of How to Use the Derivative
To find the equation of the tangent line on the curve y = Ln(x) at the point x = 2, we need to identify one point on the line and figure out the slope of the line. See image at the top of the article for a picture of the tangent line on y = Ln(x) at x = 2.
We know that the point (2, Ln(2)) lies on the tangent line, since this is where the tangent line touches the curve y = Ln(2).
From calculus, we know that the slope of the tangent line at x is equal to 1/x. Since we have x = 2, our slope is 1/2.
Now we can compute the equation of the tangent line using the point-slope formula
(y - y0) = m(x - x0)
where m is the slope and (x0, y0) is any point on the line. This gives us the tangent line equation
(y - Ln(2)) = (1/2)(x - 2)
y = (1/2)(x - 2) + Ln(2)
y = (1/2)x - 1 + Ln(2)
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