# Logarithmic Differentiation

Taking logarithms and applying the Laws of Logarithms can simplify the differentiation of complex functions. This is called Logarithmic Differentiation.

Before beginning our discussion, let's review the Laws of Logarithms.

LAWS OF LOGARITHMS:

If x and y are positive numbers, then

Law 1:

Law 2:

Law 3: If $lo{g}_{a}\left({x}^{r}\right)=rlo{g}_{a}x$

Let's look at some examples:

To work these examples requires the use of various differentiation rules. If you are not familiar with a rule go to the associated topic for a review.

$y=\frac{\sqrt[3]{{x}^{2}+3}}{{\left(x-1\right)}^{4}}$
 Step 1: Take the natural log of both sides of the equation and use the law of logarithms to simplify. Take ln of both sides. ${\mathrm{ln}}y={\mathrm{ln}}\frac{\sqrt[3]{{x}^{2}+3}}{{\left(x-1\right)}^{4}}$ Apply Laws of Logarithms. Step 2: Differentiate implicitly. To differentiate the two terms on the right-hand side, apply the Chain Rule $\frac{\partial }{\partial x}\mathrm{ln}y=\frac{\partial }{\partial x}\frac{1}{3}\mathrm{ln}\left({x}^{2}+3\right)-\frac{\partial }{\partial x}4\mathrm{ln}\left(x-1\right)$ Step 3: Solve for $\frac{\partial y}{\partial x}$. Multiply both sides by y. Step 4: Substitute for y. $y=\frac{\sqrt[3]{{x}^{2}+3}}{{\left(x-1\right)}^{4}}$
Example 1:      $y={\left(3x\right)}^{\sqrt{x}}$
 Step 1: Take the natural log of both sides of the equation and use the law of logarithms to simplify. ${\mathrm{ln}}y={\mathrm{ln}}{\left(3x\right)}^{\sqrt{x}}$ Take ln Apply Law 1 of Logarithms Step 2: Differentiate implicitly. Apply the Product & Chain Rules to the right hand side. Step 3: Solve for $\frac{\partial y}{\partial x}$ and simplify. Multiply both sides by y. $\frac{\partial y}{\partial x}=y\left[\frac{1}{\sqrt{x}}+\frac{\mathrm{ln}3x}{2\sqrt{x}}\right]$ Step 4: Substitute for y. $y={\left(3x\right)}^{\sqrt{x}}$ $\frac{\partial y}{\partial x}=\left[{\left(3x\right)}^{\sqrt{x}}\right]\left[\frac{2+\mathrm{ln}3x}{2\sqrt{x}}\right]$
Example 2:      $y={e}^{\left(5{x}^{2}-x+4\right)}$
 Step 1: Take the natural log of both sides of the equation and use the law of logarithms to simplify. Remember that . ${\mathrm{ln}}y={\mathrm{ln}}{e}^{\left(5{x}^{2}-x+4\right)}$ Take ln ${\mathrm{ln}}y=\left(5{x}^{2}-x+4\right)$ Step 2: Differentiate implicitly. $\frac{\partial }{\partial x}\mathrm{ln}y=\frac{\partial }{\partial x}\left(5{x}^{2}-x+4\right)$ Step 3: Solve for $\frac{\partial y}{\partial x}$. Multiply both sides by y. $\frac{\partial y}{\partial x}=y\left(10x-1\right)$ Step 4: Substitute for y. $y={e}^{\left(5{x}^{2}-x+4\right)}$ $\frac{\partial y}{\partial x}=\left[{e}^{\left(5{x}^{2}-x+4\right)}\right]\left(10x-1\right)$ This function can also be solved explicitly using the chain rule. Step 1: Identify the inner function and rewrite the outer function replacing the inner function by the variable u. g = 5x2 - x + 4     Inner Function u = 5x2 - x + 4 f = eu     Outer Function Step 2: Take the derivative of both functions. Derivative of f = eu $\frac{d}{dx}{e}^{u}$      Original eu          Nat. Exp. Rule ${e}^{u}$ __________________________ Derivative of g = 5x2 - x + 4 $\frac{d}{dx}5{x}^{2}-x+4$ Original $\frac{d}{dx}5{x}^{2}-\frac{d}{dx}x+\frac{d}{dx}4$    Sum/Diff Rule $5\frac{d}{dx}{x}^{2}-\frac{d}{dx}x+\frac{d}{dx}4$    Constant Multiple 5(2x) - 1 + 0 Power & Constant $10x-1$ Step 3: Substitute the derivatives and the original expression for the variable u into the Chain Rule and simplify. $\left(f\circ g\right)\left(x\right)={f}^{\prime }\left(g\left(x\right)\right)·g\prime \left(x\right)$ ${e}^{u}\left(10x-1\right)$ Chain Rule ${e}^{\left(5{x}^{2}-x+4\right)}\left(10x-1\right)$  Substitute for u ${e}^{\left(5{x}^{2}-x+4\right)}\left(10x-1\right)$

 Related Links: Math algebra Implicit Differentiation Combination of Functions Calculus Topics

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