Logarithmic Differentiation — Question 7

Question 7 - Solution

We are given: $$y = \left( \frac{(2x^2 + 1)^5}{\sqrt{5x - 3} \cdot (x^2 + 4)^3} \right)^2$$

Take the natural logarithm of both sides: $$\ln y = \ln \left[ \left( \frac{(2x^2 + 1)^5}{(5x - 3)^{1/2} \cdot (x^2 + 4)^3} \right)^2 \right]$$

Bring the exponent outside: $$\ln y = 2 \cdot \ln \left( \frac{(2x^2 + 1)^5}{(5x - 3)^{1/2} \cdot (x^2 + 4)^3} \right)$$

Use log rules: $$\ln y = 2 \left[ 5 \ln(2x^2 + 1) - \frac{1}{2} \ln(5x - 3) - 3 \ln(x^2 + 4) \right]$$

Differentiate both sides: $$\frac{1}{y} \cdot \frac{dy}{dx} = 2 \left[ \frac{5 \cdot 4x}{2x^2 + 1} - \frac{1}{2} \cdot \frac{5}{5x - 3} - \frac{3 \cdot 2x}{x^2 + 4} \right]$$

Simplify: $$\frac{dy}{dx} = y \cdot \left( \frac{40x}{2x^2 + 1} - \frac{5}{5x - 3} - \frac{12x}{x^2 + 4} \right)$$

Substitute $y$ back: $$\frac{dy}{dx} = \left( \frac{(2x^2 + 1)^5}{\sqrt{5x - 3} \cdot (x^2 + 4)^3} \right)^2 \cdot \left( \frac{40x}{2x^2 + 1} - \frac{5}{5x - 3} - \frac{12x}{x^2 + 4} \right)$$

Final Answer: $$\boxed{ \frac{dy}{dx} = \left( \frac{(2x^2 + 1)^5}{\sqrt{5x - 3} \cdot (x^2 + 4)^3} \right)^2 \cdot \left( \frac{40x}{2x^2 + 1} - \frac{5}{5x - 3} - \frac{12x}{x^2 + 4} \right) }$$