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Q.
If $y=y(x)$ follows the relation $x^2 y+3 x y^3=x-y+3$, then $y^{\prime}(0)$ equals
Continuity and Differentiability
Solution:
Given, $ x^2 y+3 x y^3=x-y+3$ .....(1)
Put$x =0 \text {, we get }$
$0=-y+3 \Rightarrow y=3$
So, $ P ( x =0, y =3)$.
Now, differentiate both sides of equation (1) with respect to $x$, we get
$x^2 \cdot \frac{d y}{d x}+2 x y+3 x\left(3 y^2\right) \frac{d y}{d x}+3 y^3 \cdot 1=1-\frac{d y}{d x}$
Put $ x=0, y=3$, we get
$ 0+0+0+81=1-\frac{ dy }{ dx } $
$\left.\therefore \frac{ dy }{ dx }\right]_{ P (0,3)}=1-81=-80$