CalcNotes Derivatives of Inverse Functions and Logarithms

Section 3.8 Derivatives of Inverse Functions and Logarithms

Example 3.8.1.

Let \(g(x)=x^2+5\text{.}\) Find \(g'(x)\text{,}\) \(g^{-1}(x)\text{,}\) and \(\frac{d}{dx} g^{-1}(x)\text{.}\)

Fact 3.8.2. Derivative Rule for Inverses.

If \(f\) has domain on an interval \(I\) and \(f^{-1}\) exists and is never 0 over \(I\text{,}\) then \(f^{-1}\) is differentiable on its domain. The value of \((f^{-1})'\) at a point \(b\) is the reciprocal of the value of \(f'\) at \(f^{-1}(b)\text{,}\) i.e.,

\begin{equation*} (f^{-1})'(b) = \frac{1}{f'(f^{-1}(b))} \end{equation*}

Subsection 3.8.1 Derivatives of Logarithms

Theorem 3.8.3.

Let \(f(x) = e^x\) so that \(f^{-1}(x) = \ln x\) for \(x\gt 0\text{.}\) Then

\begin{equation*} \frac{d}{dx} \ln x = \frac{1}{x} \end{equation*}

Proof.

The following relies on content from Section 3.8, but is included here for reference.

\begin{equation*} \frac{d}{dx} \ln x = (f^{-1}(x))' = \frac{1}{f'(f^{-1}(x))} = \frac{1}{e^{f^{-1}(x)}} = \frac{1}{e^{\ln x}} = \frac{1}{x} \end{equation*}

Fact 3.8.4. Derivative of \(ln x\).

Combining with the chain rule produces:

\begin{equation*} \frac{d}{dx} \ln u = \frac{1}{u} \frac{du}{dx} \end{equation*}

Example 3.8.5.

Compute the derivatives:

\(\displaystyle \frac{d}{dx} 5 \ln (7x)\)
\(\displaystyle \frac{d}{dx} \ln |x|\)
Let \(y=7 \ln (5\ln x)\text{.}\) Find \(y'\text{.}\)

Answer.

\(\displaystyle \displaystyle \frac{5}{x}\)
\(\displaystyle \displaystyle \frac{1}{x}\)
\(\displaystyle \displaystyle \frac{7}{x\ln x}\)

Subsection 3.8.2 Derivatives of \(a^u\) and \(\log_a u\)

Theorem 3.8.6.

\begin{equation*} \frac{d}{dx} a^u = (\ln a) a^u \frac{du}{dx} \end{equation*}

\begin{equation*} \frac{d}{dx} \log_a u = \frac{1}{(\ln a)u} \frac{du}{dx} \end{equation*}

Proof.

\begin{align*} \frac{d}{dx} a^u \amp= \frac{d}{dx} e^{\ln a^u}\\ \amp= \frac{d}{dx} e^{u \ln a}\\ \amp= e^{u \ln a} \frac{d}{dx} u \ln a\\ \amp= \ln a e^{u \ln a} \frac{du}{dx}\\ \amp= \ln a a^u \frac{du}{dx}\\ \frac{d}{dx} \log_a u \amp= \frac{d}{dx} \frac{\ln u}{\ln a}\\ \amp= \frac{1}{u\ln a} \frac{du}{dx} \end{align*}

Subsection 3.8.3 Logarithmic differentiation

Example 3.8.7.

Let \(y=\ln(8\cos^3(x))\text{.}\) Find \(y'\text{.}\)

Answer.

\(\displaystyle y' = -3\tan x\)

Taking the natural log of both sides can make it easier to differentiate.

Example 3.8.8.

Let \(y=\sqrt{x(x+10)}\text{.}\) Find \(\frac{dy}{dx}\text{.}\)
Let \(y=\sqrt{\theta+4}\sin(9\theta)\text{.}\) Find \(y'\text{.}\)

Answer.

\(\displaystyle \displaystyle \frac{dy}{dx} = \frac12 \sqrt{x(x+10)} \left(\frac{1}{x} + \frac{1}{x+10}\right)\)
\(\displaystyle \displaystyle y' = \sqrt{\theta + 4} \left(\frac{\sin (9\theta)}{2\theta + 8}+9\cos (9\theta)\right)\)

Note that logarithmic differentiation assumes that the derivative exists.

Example 3.8.9.

On the domain \(x\gt 0\text{,}\) find \(\frac{d}{dx} x^x\text{.}\)

Let \(y = (9x+7)^x\text{;}\) find \(y'\text{.}\)

Answer.

\(\displaystyle x^x(\ln x + 1)\)
\(\displaystyle \displaystyle (9x+7)^x \left(\ln(9x+7)+\frac{9x}{9x+7}\right)\)

Definition 3.8.10. Definition of \(e\).

\begin{equation*} e = \lim_{x\rightarrow 0} (1+x)^{1/x} \end{equation*}