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Differentiation (p-1) — Maths (commerce) STD 12 Commerce / Arts — Question

Maharashtra BoardEnglish MediumSTD 12 Commerce / ArtsMaths (commerce)Differentiation (p-1)2 Marks
Question
Find $\frac{d y}{d x}$ if, :
$
x^3+y^3+4 x^3 y=0
$
✓

Answer

$
x^3+y^3+4 x^3 y=0
$
Differentiating both sides w.r.t. $x_r$ we get
$
\begin{aligned}
& 3 x^2+3 y^2 \frac{d y}{d x}+4\left[x^3 \frac{d y}{d x}+y \frac{d}{d x}\left(x^3\right)\right]=0 \\
& \therefore 3 x^2+3 y^2 \frac{d y}{d x}+4 x^3 \frac{d y}{d x}+4 y \times 3 x^2=0 \\
& \therefore 3 y^2 \frac{d y}{d x}+4 x^3 \frac{d y}{d x}=-3 x^2-12 x^2 y \\
& \therefore\left(3 y^2+4 x^3\right) \frac{d y}{d x}=-3 x^2(1+4 y) \\
& \therefore \frac{d y}{d x}=-\frac{3 x^2(1+4 y)}{3 y^2+4 x^3} .
\end{aligned}
$

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