x3y−2×2+y4=8 then dydx at P(1,2) is

# ${x}^{3}y-2{x}^{2}+{y}^{4}=8$ then $\frac{dy}{dx}$ at $P\left(1,2\right)$ is

1. A

$\frac{2}{33}$

2. B

$\frac{-2}{33}$

3. C

$\frac{4}{15}$

4. D

$\frac{-4}{15}$

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### Solution:

$\begin{array}{l}{x}^{3}y-2{x}^{2}+{y}^{4}=8\\ ⇒{x}^{3}\frac{dy}{dx}+y\left(3{x}^{2}\right)-2\left(2x\right)+4{y}^{3}\frac{dy}{dx}=0\\ ⇒{x}^{3}\frac{dy}{dx}+4{y}^{3}\frac{dy}{dx}=4x-3{x}^{2}y\\ ⇒\left({x}^{3}+4{y}^{3}\right)\frac{dy}{dx}=4x-3{x}^{2}y\\ ⇒\frac{dy}{dx}=\frac{4x-3{x}^{2}y}{{x}^{3}+4{y}^{3}}\\ ⇒{\left(\frac{dy}{dx}\right)}_{P\left(1,2\right)}=\frac{4\left(1\right)-3{\left(1\right)}^{2}\left(2\right)}{{\left(1\right)}^{3}+4{\left(2\right)}^{3}}\\ =\frac{4-6}{1+32}\\ =\frac{-2}{33}\end{array}$

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