Find d y d x if, : x^3+x^2 y+x y^2+y^3=81 - Vidyadip

Question

Find $\frac{d y}{d x}$ if, :
$
x^3+x^2 y+x y^2+y^3=81
$

✓

Answer

$
x^3+x^2 y+x y^2+y^3=81
$
Differentiating both sides w.r.t. $x_1$ we get
$
\begin{array}{r}
3 x^2+\left[x^2 \frac{d y}{d x}+y \cdot \frac{d}{d y}\left(x^2\right)\right]+\left[x \cdot \frac{d}{d x}\left(y^2\right)+y^2 \cdot \frac{d}{d x}(x)\right]+ \\
3 y^2 \frac{d y}{d x}=0
\end{array}
$
$
\therefore 3 x^2+x^2 \frac{d y}{d x}+y \times 2 x+x \times 2 y \cdot \frac{d y}{d x}+y^2 \times 1+
$
$
3 y^2 \frac{d y}{d x}=0
$
$
\begin{aligned}
& \therefore x^2 \frac{d y}{d x}+2 x y \frac{d y}{d x}+3 y^2 \frac{d y}{d x}=-3 x^2-2 x y-y^2 \\
& \therefore\left(x^2+2 x y+3 y^2\right) \frac{d y}{d x}=-\left(3 x^2+2 x y+y^2\right) \\
& \therefore \frac{d y}{d x}=-\left(\frac{3 x^2+2 x y+y^2}{x^2+2 x y+3 y^2}\right) .
\end{aligned}
$

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