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If $x^2y +yx^3=5x+3 ,$ then $\frac{dy}{dx}= $ |
$\frac{5}{2x+3x^2-3}$ $-\frac{3x^2y-2xy}{x^2+x^3-3}$ $\frac{5-3x^2y-3x}{x^2+x^3-3}$ $\frac{5-3x^2y-2xy}{x^2+x^3-3}$ |
$\frac{5-3x^2y-2xy}{x^2+x^3-3}$ |
The correct answer is Option (4) → $\frac{5-3x^2y-2xy}{x^2+x^3-3}$ $x^2y +yx^3=5x+3$ Differentiating both sides w.r.t. 'x'. $2xy+x^2\frac{dy}{dx}+3x^2y+x^3\frac{dy}{dx}=5$ $\frac{dy}{dx}(x^2+x^3)=5-2xy-3x^2y$ $⇒\frac{dy}{dx}=\frac{5-2xy-3x^2y}{x^2+x^3}$ |
