Derivative of the Logarithmic Function

Proof

If \(x > 0\) and \(y = \text{ln} x ,\) then \(e^{y} = x .\) Differentiating both sides of this equation results in the equation

Solving for \(\frac{d y}{d x}\) yields

Finally, we substitute \(x = e^{y}\) to obtain

We may also derive this result by applying the inverse function theorem, as follows. Since \(y = g \left(\right. x \left.\right) = \text{ln} x\) is the inverse of \(f \left(\right. x \left.\right) = e^{x} ,\) by applying the inverse function theorem we have

Using this result and applying the chain rule to \(h \left(\right. x \left.\right) = \text{ln} \left(\right. g \left(\right. x \left.\right) \left.\right)\) yields

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The graph of \(y = \text{ln} x\) and its derivative \(\frac{d y}{d x} = \frac{1}{x}\) are shown in Figure 3.35.

Figure 3.35 \(\text{The function} y = \text{ln} x\) is increasing on \(\left(\right. 0 , + \infty \left.\right) .\) Its derivative \(y ' = \frac{1}{x}\) is greater than zero on \(\left(\right. 0 , + \infty \left.\right) .\)

Now that we can differentiate the natural logarithmic function, we can use this result to find the derivatives of \(y = l o g_{b} x\) and \(y = b^{x}\) for \(b > 0 , b \neq 1 .\)

Proof

If \(y = \text{log}_{b} x ,\) then \(b^{y} = x .\) It follows that \(\text{ln} \left(\right. b^{y} \left.\right) = \text{ln} x .\) Thus \(y \text{ln} b = \text{ln} x .\) Solving for \(y ,\) we have \(y = \frac{\text{ln} x}{\text{ln} b} .\) Differentiating and keeping in mind that \(\text{ln} b\) is a constant, we see that

The derivative in Equation 3.32 now follows from the chain rule.

If \(y = b^{x} ,\) then \(\text{ln} y = x \text{ln} b .\) Using implicit differentiation, again keeping in mind that \(\text{ln} b\) is constant, it follows that \(\frac{1}{y} \frac{d y}{d x} = \text{ln} b .\) Solving for \(\frac{d y}{d x}\) and substituting \(y = b^{x} ,\) we see that

The more general derivative (Equation 3.35) follows from the chain rule.

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