Find :
Question:

Find $\frac{d y}{d x}$

$x^{2}+x y+y^{2}=100$

Solution:

The given relationship is $x^{2}+x y+y^{2}=100$

Differentiating this relationship with respect to x, we obtain

$\frac{d}{d x}\left(x^{2}+x y+y^{2}\right)=\frac{d}{d x}(100)$

$\Rightarrow \frac{d}{d x}\left(x^{2}\right)+\frac{d}{d x}(x y)+\frac{d}{d x}\left(y^{2}\right)=0$  [Derivative of constant function is 0 ]

$\Rightarrow 2 x+\left[y \cdot \frac{d}{d x}(x)+x \cdot \frac{d y}{d x}\right]+2 y \frac{d y}{d x}=0$   [Using product rule and chain rule]

$\Rightarrow 2 x+y \cdot 1+x \cdot \frac{d y}{d x}+2 y \frac{d y}{d x}=0$

$\Rightarrow 2 x+y+(x+2 y) \frac{d y}{d x}=0$

$\therefore \frac{d y}{d x}=-\frac{2 x+y}{x+2 y}$

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