Find d y d x if : x^5+x y^3+x^2 y+y^4=4 - Vidyadip

Question

Find $\frac{d y}{d x}$ if : $x^5+x y^3+x^2 y+y^4=4$

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Answer

Given that: $x^5+x y^3+x^2 y+y^4=4$
Differentiate $w, r, t, x$.
$
\begin{aligned}
& \frac{d}{d x}\left(x^5\right)+\frac{d}{d x}\left(x y^3\right)+\frac{d}{d x}\left(x^2 y\right)+\frac{d}{d x}\left(y^4\right)=\frac{d}{d x}(4) \\
& 5 x^4+x \frac{d}{d x}\left(y^3\right)+y^3 \frac{d}{d x}(x)+x^2 \frac{d}{d x}(y)+y \frac{d}{d x}\left(x^2\right)+4 y^3 \frac{d}{d x}(y)=0 \\
& 5 x^4+x\left(3 y^2\right) \frac{d y}{d x}+y^3(1)+x^2 \frac{d y}{d x}+y(2 x)+4 y^3 \frac{d y}{d x}=0 \\
& x^2 \frac{d y}{d x}+3 x y^2 \frac{d y}{d x}+4 y^3 \frac{d y}{d x}=-5 x^4-2 x y-y^3 \\
& \left(x^2+3 x y^2+4 y^3\right) \frac{d y}{d x}=-\left(5 x^4+2 x y+y^3\right) \\
\therefore \quad & \frac{d y}{d x}=-\frac{5 x^4+2 x y+y^3}{x^2+3 x y^2+4 y^3}
\end{aligned}
$

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