Ulearngo

Logarithmic Differentiation

At this point, we can take derivatives of functions of the form \(y = \left(\right. g \left(\right. x \left.\right) \left.\right)^{n}\) for certain values of \(n ,\) as well as functions of the form \(y = b^{g \left(\right. x \left.\right)} ,\) where \(b > 0\) and \(b \neq 1 .\) Unfortunately, we still do not know the derivatives of functions such as \(y = x^{x}\) or \(y = x^{\pi} .\) These functions require a technique called logarithmic differentiation, which allows us to differentiate any function of the form \(h \left(\right. x \left.\right) = g \left(\right. x \left.\right)^{f \left(\right. x \left.\right)} .\) It can also be used to convert a very complex differentiation problem into a simpler one, such as finding the derivative of \(y = \frac{x \sqrt{2 x + 1}}{e^{x} \text{sin}^{3} x} .\) We outline this technique in the following problem-solving strategy.

Problem-Solving Strategy

Using Logarithmic Differentiation

  1. To differentiate \(y = h \left(\right. x \left.\right)\) using logarithmic differentiation, take the natural logarithm of both sides of the equation to obtain \(\text{ln} y = \text{ln} \left(\right. h \left(\right. x \left.\right) \left.\right) .\)
  2. Use properties of logarithms to expand \(\text{ln} \left(\right. h \left(\right. x \left.\right) \left.\right)\) as much as possible.
  3. Differentiate both sides of the equation. On the left we will have \(\frac{1}{y} \frac{d y}{d x} .\)
  4. Multiply both sides of the equation by \(y\) to solve for \(\frac{d y}{d x} .\)
  5. Replace \(y\) by \(h \left(\right. x \left.\right) .\)

Example 3.81

Using Logarithmic Differentiation

Find the derivative of \(y = \left(\right. 2 x^{4} + 1 \left.\right)^{\text{tan} x} .\)

Solution

Use logarithmic differentiation to find this derivative.

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

Example 3.82

Using Logarithmic Differentiation

Find the derivative of \(y = \frac{x \sqrt{2 x + 1}}{e^{x} \text{sin}^{3} x} .\)

Solution

This problem really makes use of the properties of logarithms and the differentiation rules given in this chapter.

\[\begin{aligned} \text{ln} y & = & \text{ln} \frac{x \sqrt{2 x + 1}}{e^{x} \text{sin}^{3} x} & & & \text{Step 1}.\text{ Take the natural logarithm of both sides}. \\ \text{ln} y & = & \text{ln} x + \frac{1}{2} \text{ln} \left(\right. 2 x + 1 \left.\right) - x \text{ln} e - 3 \text{ln} \text{sin} x & & & \text{Step 2}.\text{ Expand using properties of logarithms}. \\ \frac{1}{y} \frac{d y}{d x} & = & \frac{1}{x} + \frac{1}{2 x + 1} - 1 - 3 \frac{\text{cos} x}{\text{sin} x} & & & \text{Step 3}.\text{ Differentiate both sides}. \\ \frac{d y}{d x} & = & y \left(\right. \frac{1}{x} + \frac{1}{2 x + 1} - 1 - 3 \text{cot} x \left.\right) & & & \text{Step 4}.\text{ Multiply by} y \text{on both sides}. \\ \frac{d y}{d x} & = & \frac{x \sqrt{2 x + 1}}{e^{x} \text{sin}^{3} x} \left(\right. \frac{1}{x} + \frac{1}{2 x + 1} - 1 - 3 \text{cot} x \left.\right) & & & \text{Step 5}.\text{ Substitute} y = \frac{x \sqrt{2 x + 1}}{e^{x} \text{sin}^{3} x} . \end{aligned}\]

Example 3.83

Extending the Power Rule

Find the derivative of \(y = x^{r}\) where \(r\) is an arbitrary real number.

Solution

The process is the same as in Example 3.82, though with fewer complications.

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

Checkpoint 3.54

Use logarithmic differentiation to find the derivative of \(y = x^{x} .\)

Checkpoint 3.55

Find the derivative of \(y = \left(\right. \text{tan} x \left.\right)^{\pi} .\)

This lesson is part of:

Derivatives

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