Logarithmic Differentiation

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.

$\ln y=\ln \frac{\sqrt[3]{x^2+3}}{{\left(x-1\right)}^4}$

Apply Laws of Logarithms.

$\ln y=\frac{1}{3}\ln \left(x^2+3\right)-4 \text{ln}\left(x-1\right)$

Step 2: Differentiate implicitly.

To differentiate the two terms on the right-hand side, apply the Chain Rule

$\frac{\partial }{\partial x}\ln y=\frac{\partial }{\partial x}\frac{1}{3}\ln \left(x^2+3\right)-\frac{\partial }{\partial x}4\ln \left(x-1\right)$

$\frac{1}{y} \frac{\partial y}{\partial x}=\left[\frac{1}{3} ·\frac{1}{x^2+3} ·2x\right]-\left[4 ·\frac{1}{x-1} ·1\right]$

$\frac{1}{y} \frac{\partial y}{\partial x}=\frac{2x}{3x^2+9} -\frac{4}{x-1}$

Step 3: Solve for $\frac{\partial y}{\partial x}$.

Multiply both sides by y.

$\frac{\partial y}{\partial x}=y\left[\frac{2x}{3x^2+9} -\frac{4}{x-1}\right]$

Step 4: Substitute for y.

$y=\frac{\sqrt[3]{x^2+3}}{{\left(x-1\right)}^4}$

$\frac{\partial y}{\partial x}=\left[\frac{\sqrt[3]{x^2+3}}{{\left(x-1\right)}^4}\right]\left[\frac{2x}{3x^2+9} -\frac{4}{x-1}\right]$

Step 1: Take the natural log of both sides of the equation and use the law of logarithms to simplify.

$\ln y=\ln {\left(3x\right)}^{\sqrt{x}}$ Take ln

Apply Law 1 of Logarithms

$\ln y=\sqrt{x}\text{ ln}\left(3x\right)$

Step 2: Differentiate implicitly.

Apply the Product & Chain Rules to the right hand side.

$\frac{\partial }{\partial x}\ln y=\frac{\partial }{\partial x}\sqrt{x}\text{ ln}\left(3x\right)$

$\frac{1}{y} \frac{\partial y}{\partial x}=\sqrt{x}·\frac{1}{3x}·3+\ln 3x ·\frac{1}{2} x^{-\frac{1}{2}}$

$\frac{1}{y} \frac{\partial y}{\partial x}=\frac{3}{3x^{\frac{3}{2}}}+\frac{\ln 3x}{2\sqrt{x}}$

Step 3: Solve for $\frac{\partial y}{\partial x}$ and simplify.

Multiply both sides by y.

$\frac{\partial y}{\partial x}=y\left[\frac{1}{\sqrt{x}}+\frac{\ln 3x}{2\sqrt{x}}\right]$

Step 4: Substitute for y.

$y={\left(3x\right)}^{\sqrt{x}}$

$\frac{\partial y}{\partial x}=\left[{\left(3x\right)}^{\sqrt{x}}\right]\left[\frac{2+\ln 3x}{2\sqrt{x}}\right]$

Step 1: Take the natural log of both sides of the equation and use the law of logarithms to simplify.

Remember that $\ln e^x=e^x \therefore \ln e^{\left(5x^2-x+4\right)}=5x^2-x+4$.

$\ln y=\ln e^{\left(5x^2-x+4\right)}$ Take ln

$\ln y=\left(5x^2-x+4\right)$

Step 2: Differentiate implicitly.

$\frac{\partial }{\partial x}\ln y=\frac{\partial }{\partial x}\left(5x^2-x+4\right)$

$\frac{1}{y} \frac{\partial y}{\partial x}=\left[5\left(2\text{x}\right)-1\right]$

$\frac{1}{y} \frac{\partial y}{\partial x}=10x-1$

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.

$y=e^{\left(5x^2-x+4\right)}$

$\frac{\partial y}{\partial x}=\left[e^{\left(5x^2-x+4\right)}\right]\left(10x-1\right)$

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 = e^u     Outer Function

Step 2: Take the derivative of both functions.

Derivative of f = e^u

$\frac{d}{dx}{e}^u$      Original

e^u          Nat. Exp. Rule

$e^u$

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Derivative of g = 5x² - x + 4

$\frac{d}{dx}5x^2-x+4$ Original

$\frac{d}{dx}5x^2-\frac{d}{dx}x+\frac{d}{dx}4$    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

$e^{\left(5x^2-x+4\right)}\left(10x-1\right)$  Substitute for u

$e^{\left(5x^2-x+4\right)}\left(10x-1\right)$