Math  /  Calculus

QuestionUse logarithmic differentiation to find the derivative. y=x2+3x2+45y=\sqrt[5]{\frac{x^{2}+3}{x^{2}+4}}

Studdy Solution
Solve for dydx \frac{dy}{dx} :
1ydydx=15(2xx2+32xx2+4) \frac{1}{y} \frac{dy}{dx} = \frac{1}{5} \left( \frac{2x}{x^{2}+3} - \frac{2x}{x^{2}+4} \right)
Multiply both sides by y y :
dydx=y15(2xx2+32xx2+4) \frac{dy}{dx} = y \cdot \frac{1}{5} \left( \frac{2x}{x^{2}+3} - \frac{2x}{x^{2}+4} \right)
Substitute back the original expression for y y :
dydx=x2+3x2+4515(2xx2+32xx2+4) \frac{dy}{dx} = \sqrt[5]{\frac{x^{2}+3}{x^{2}+4}} \cdot \frac{1}{5} \left( \frac{2x}{x^{2}+3} - \frac{2x}{x^{2}+4} \right)
The derivative of y y is:
dydx=15x2+3x2+45(2xx2+32xx2+4) \frac{dy}{dx} = \frac{1}{5} \sqrt[5]{\frac{x^{2}+3}{x^{2}+4}} \left( \frac{2x}{x^{2}+3} - \frac{2x}{x^{2}+4} \right)

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