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Find the total differential coefficient of x^2y with respect to x when x, y are related by x^2+y^2+xy=1

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\text{Given the constraint: } x^2 + y^2 + xy = 1

\text{Differentiate implicitly w.r.t. } x:

$2x + 2y\,\frac{dy}{dx} + y + x\,\frac{dy}{dx} = 0$

\text{Collect the } \frac{dy}{dx} \text{ terms:}

$(2y + x)\frac{dy}{dx} = -(2x + y)$

\text{Thus,}

$\frac{dy}{dx} = -\,\frac{2x + y}{2y + x}$

\text{Now consider } f = x^2 y.

\text{Differentiate:}

$\frac{df}{dx} = \frac{d}{dx}(x^2 y) = 2xy + x^2 \frac{dy}{dx}$

\text{Substitute } \frac{dy}{dx}: 

$\frac{df}{dx} = 2xy + x^2\left( -\,\frac{2x + y}{2y + x} \right)$

\text{Therefore, the total differential coefficient is:}

$\boxed{\frac{df}{dx} = 2xy - \frac{x^2(2x + y)}{2y + x}}$
 

Posted by

Saumya Singh

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