Logarithmic Differentiation

Logarithmic differentiation is a powerful method for differentiating functions that are difficult to handle using standard rules. It is especially useful when:

the variable appears in both the base and the exponent
the function is a product of many terms
the function is a quotient of complicated expressions

Refresher of log laws and indices

Example

Differentiate:

\[ y = 5^{2x} \]

Step 1 — Take logs of both sides

\[ \ln y = \ln\left(5^{2x}\right) = 2x \ln 5. \]

Step 2 — Differentiate differentiate both sides implicitly

reminder on differentiating logarithms

\[ \frac{1}{y}\frac{dy}{dx} = 2\ln 5. \]

Step 3 — Substitute back for \(y\)

\[ \frac{dy}{dx} = 2(\ln 5)\,y = 5^{2x}(2\ln 5) = 5^{2x}(\ln 5^{2}) = 25^{x}(\ln 25) \]

Example

Differentiate:

\[ y = \frac{(x^2 + 1)^5}{\sqrt{x^3 - 4x}} \]

Step 1 — Take logs and apply rules for division

\[ \ln y = 5\ln(x^2 + 1) - \tfrac{1}{2}\ln(x^3 - 4x). \]

Step 2 — Differentiate implicitly

\[ \frac{1}{y}\frac{dy}{dx} = 5\left(\frac{2x}{x^2 + 1}\right) - \tfrac{1}{2}\left(\frac{3x^2 - 4}{x^3 - 4x}\right). \]

Step 3 — Multiply both sides by \(y\)

\[ \frac{dy}{dx} = \frac{(x^2 + 1)^5}{\sqrt{x^3 - 4x}} \left[ \frac{10x}{x^2 + 1} - \frac{3x^2 - 4}{2(x^3 - 4x)} \right]. \]

\[ \frac{dy}{dx} = \frac{(x^2 + 1)^5}{\sqrt{x^3 - 4x}} \left[ \frac{10x}{x^2 + 1} - \frac{3x^2 - 4}{2(x^3 - 4x)} \right] \]

\[ \text{Simplify the bracket by using a common denominator } 2(x^2+1)(x^3-4x): \] \[ \frac{10x}{x^2+1} = \frac{20x(x^3-4x)}{2(x^2+1)(x^3-4x)}, \quad \frac{3x^2 - 4}{2(x^3 - 4x)} = \frac{(3x^2-4)(x^2+1)}{2(x^2+1)(x^3-4x)}. \] \[ \Rightarrow \left[ \frac{10x}{x^2 + 1} - \frac{3x^2 - 4}{2(x^3 - 4x)} \right] = \frac{20x(x^3-4x) - (3x^2-4)(x^2+1)}{2(x^2+1)(x^3-4x)}. \]

\[ 20x(x^3-4x) = 20x^4 - 80x^2, \] \[ (3x^2-4)(x^2+1) = 3x^4 - x^2 - 4, \] \[ \Rightarrow 20x^4 - 80x^2 - (3x^4 - x^2 - 4) = 17x^4 - 79x^2 + 4. \] \[ \therefore \left[ \frac{10x}{x^2 + 1} - \frac{3x^2 - 4}{2(x^3 - 4x)} \right] = \frac{17x^4 - 79x^2 + 4}{2(x^2+1)(x^3-4x)}. \]

\[ \frac{dy}{dx} = \frac{(x^2 + 1)^5}{\sqrt{x^3 - 4x}} \cdot \frac{17x^4 - 79x^2 + 4}{2(x^2+1)(x^3-4x)} \] \[ = \frac{(x^2+1)^4(17x^4 - 79x^2 + 4)}{2(x^3-4x)\sqrt{x^3-4x}} = \frac{(x^2+1)^4(17x^4 - 79x^2 + 4)}{2(x^3-4x)^{3/2}}. \]