Question

If $x^4+7x^2{y}^2+9y^4=24x{y}^3$, then $\frac{dy}{dx}=$

• A. $\frac{x}{y}$
• B. $\frac{y}{x}$
• C. $-\frac{x}{y}$
• D. $-\frac{y}{x}$

Answer 1

Answer: (B)

$\frac{y}{x}$

$x^4+7x^2{y}^2+9y^4=24x{y}^3$

$4x^3+14x{y}^2+14x^{2}y\frac{dy}{dx}+36y^3\frac{dy}{dx}=24y^3+72x{y}^2\frac{dy}{dx}$

$\frac{dy}{dx}(14x^{2}y+36y^3-72x{y}^2)=24y^3-4x^3-14x{y}^2$

$\frac{dy}{dx}=\frac{24y^3-4x^3-14x{y}^2}{14x^{2}y+36y^3-72x{y}^2}$

$\frac{dy}{dx}=\frac{y}{x}\cdot \frac{x}{y}\cdot \frac{24y^3-4x^3-14x{y}^2}{14x^{2}y+36y^3-72x{y}^2}$

$\frac{dy}{dx}=\frac{y}{x}\cdot \frac{24x{y}^3-4x^4-14x^2{y}^2}{14x^2{y}^2+36y^4-72x{y}^3}$

Substitute $x^4=24x{y}^3-7x^2{y}^2-9y^4$

$\frac{dy}{dx}=\frac{y}{x}\cdot \frac{24x{y}^3-4(24x{y}^3-7x^2{y}^2-9y^4)-14x^2{y}^2}{14x^2{y}^2+36y^4-72x{y}^3}$

$\frac{dy}{dx}=\frac{y}{x}\cdot \left(\frac{14x^2{y}^2+36y^4-72x{y}^3}{14x^2{y}^2+36y^4-72x{y}^3}\right)$

$\therefore \frac{dy}{dx}=\frac{y}{x}$