Question

If $y=y(x)$ follows the relation $x^{2}y+3x{y}^3=x-y+3$, then $y^{\prime }(0)$ equals

• A. -27
• B. -54
• C. -80
• D. -100

Answer 1

Answer: (C)

Given, $x^{2}y+3x{y}^3=x-y+3$ .....(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
$x^2\cdot \frac{dy}{dx}+2xy+3x\left(3y^2\right)\frac{dy}{dx}+3y^3\cdot 1=1-\frac{dy}{dx}$
Put $x=0,y=3$, we get
$0+0+0+81=1-\frac{dy}{dx}$
${\therefore \frac{dy}{dx}]}_{P(0,3)}=1-81=-80$