Derivatives of Exponential and Logarithmic Functions: Learn It 3

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Derivatives of Exponential and Logarithmic Functions: Learn It 3

Logarithmic Differentiation

We now explore derivatives of functions like $y=(g(x))^n$ for certain values of $n$ , and those involving forms like $y=b^{g(x)}$ , where $b>0$ and $b\ne 1$ . These functions present a challenge when trying to find derivatives directly. To address this, we employ logarithmic differentiation. This technique is particularly useful for differentiating functions of the form $y=x^x$ or $y=x^{\pi}$ .

Logarithmic differentiation simplifies the process by taking the natural logarithm of both sides, which transforms multiplicative relationships into additive ones, making the derivative more straightforward to compute. We will apply this technique to solve problems such as finding the derivate of $y=\frac{x\sqrt{2x+1}}{e^x \sin^3 x}$ .

We outline this technique in the following problem-solving strategy.

How To: Use Logarithmic Differentiation

Begin by Logarithmizing: To apply logarithmic differentiation to the function $y=h(x)$ , start by taking the natural logarithm of both sides of the equation, giving $\ln y=\ln (h(x))$ .
Expand Using Logarithm Properties: Utilize the properties of logarithms to simplify $\ln (h(x))$ as much as possible. This may involve using the logarithm rules for multiplication, division, and powers.
Differentiate Both Sides: With $y$ implicitly defined by the equation $\ln y=\ln h(x)$ , differentiate both sides with respect to $x$ . On the left, apply the chain rule to obtain $\frac{1}{y}\frac{dy}{dx}$ , and on the right, use the derivative of $\ln h(x)$ .
Isolate $\frac{dy}{dx}$ : Multiply through by $y$ to solve for $\frac{dy}{dx}$ .
Simplify the Derivative: Simplify the expression for $\frac{dy}{dx}$ to obtain the final derivative in terms of $x$ .

It may be useful to review your properties of logarithms. These will help us in step 2 to expand our logarithmic function.

The Product Rule for Logarithms	${\mathrm{log}}_{b}\left(MN\right)={\mathrm{log}}_{b}\left(M\right)+{\mathrm{log}}_{b}\left(N\right)$
The Quotient Rule for Logarithms	${\mathrm{log}}_{b}\left(\frac{M}{N}\right)={\mathrm{log}}_{b}M-{\mathrm{log}}_{b}N$
The Power Rule for Logarithms	${\mathrm{log}}_{b}\left({M}^{n}\right)=n{\mathrm{log}}_{b}M$
The Change-of-Base Formula	${\mathrm{log}}_{b}M\text{=}\frac{{\mathrm{log}}_{n}M}{{\mathrm{log}}_{n}b}\text{ }n>0,n\ne 1,b\ne 1$

Find the derivative of $y=(2x^4+1)^{\tan x}$

Use logarithmic differentiation to find this derivative.

\begin{array}{lllll} \ln y & = \ln (2 x^{4} + 1)^{\tan x} & Step 1. Take the natural logarithm of both sides. \\ \ln y & = \tan x \ln (2 x^{4} + 1) & Step 2. Expand using properties of logarithms. \\ \frac{1}{y} \frac{d y}{d x} & = \sec^{2} x \ln (2 x^{4} + 1) + \frac{8 x^{3}}{2 x^{4} + 1} \cdot \tan x & \begin{array}{l} Step 3. Differentiate both sides. Use the \\ product rule on the right. \end{array} \\ \frac{d y}{d x} & = y \cdot (\sec^{2} x \ln (2 x^{4} + 1) + \frac{8 x^{3}}{2 x^{4} + 1} \cdot \tan x) & Step 4. Multiply by y on both sides. \\ \frac{d y}{d x} & = (2 x^{4} + 1)^{\tan x} (\sec^{2} x \ln (2 x^{4} + 1) + \frac{8 x^{3}}{2 x^{4} + 1} \cdot \tan x) & Step 5. Substitute y = (2 x^{4} + 1)^{\tan x} . \end{array}

$\begin{array}{lllll} \ln y & = \ln (2x^4+1)^{\tan x} & & & \text{Step 1. Take the natural logarithm of both sides.} \\ \ln y & = \tan x \ln (2x^4+1) & & & \text{Step 2. Expand using properties of logarithms.} \\ \frac{1}{y}\frac{dy}{dx} & = \sec^2 x \ln (2x^4+1)+\frac{8x^3}{2x^4+1} \cdot \tan x & & & \begin{array}{l}\text{Step 3. Differentiate both sides. Use the} \\ \text{product rule on the right.} \end{array} \\ \frac{dy}{dx} & =y \cdot (\sec^2 x \ln (2x^4+1)+\frac{8x^3}{2x^4+1} \cdot \tan x) & & & \text{Step 4. Multiply by} \, y \, \text{on both sides.} \\ \frac{dy}{dx} & = (2x^4+1)^{\tan x}(\sec^2 x \ln (2x^4+1)+\frac{8x^3}{2x^4+1} \cdot \tan x) & & & \text{Step 5. Substitute} \, y=(2x^4+1)^{\tan x}.\end{array}$

Watch the following video to see the worked solution to this example.

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You can view the transcript for this segmented clip of “3.9 Derivatives of Exponential and Logarithmic Functions” here (opens in new window).

Find the derivative of $y=x^r$ where $r$ is an arbitrary real number.

The process is the same as in the last example, though with fewer complications.

\begin{array}{lllll} \ln y & = \ln x^{r} & Step 1. Take the natural logarithm of both sides. \\ \ln y & = r \ln x & Step 2. Expand using properties of logarithms. \\ \frac{1}{y} \frac{d y}{d x} & = r \frac{1}{x} & Step 3. Differentiate both sides. \\ \frac{d y}{d x} & = y \frac{r}{x} & Step 4. Multiply by y on both sides. \\ \frac{d y}{d x} & = x^{r} \frac{r}{x} & Step 5. Substitute y = x^{r} . \\ \frac{d y}{d x} & = r x^{r - 1} & Simplify. \end{array}

$\begin{array}{lllll} \ln y & = \ln x^r & & & \text{Step 1. Take the natural logarithm of both sides.} \\ \ln y & = r \ln x & & & \text{Step 2. Expand using properties of logarithms.} \\ \frac{1}{y}\frac{dy}{dx} & = r \frac{1}{x} & & & \text{Step 3. Differentiate both sides.} \\ \frac{dy}{dx} & = y \frac{r}{x} & & & \text{Step 4. Multiply by} \, y \, \text{on both sides.} \\ \frac{dy}{dx} & = x^r \frac{r}{x} & & & \text{Step 5. Substitute} \, y=x^r. \\ \frac{dy}{dx} & = rx^{r-1} & & & \text{Simplify.} \end{array}$