Question

If $x^2+y^2=4$, then the value of $\frac{dy}{dx}$ at the point $(0,2)$ is:

• A. $0$
• B. $-32$
• C. $4$
• D. $-2$

Answer 1

The correct option is A $0$

$x^2+y^2=4$

Differentiating both sides w.r.t. $x,$ we get:

$\Rightarrow$ $2x+2y\frac{dy}{dx}=0$

$\Rightarrow$ $2y\frac{dy}{dx}=-2x$

$\Rightarrow$ $\frac{dy}{dx}=\frac{-2x}{2y}$

$\Rightarrow$ $\frac{dy}{dx}=\frac{-x}{y}$

$\Rightarrow$ ${\left(\frac{dy}{dx}\right)}_{(0,2)}=\frac{0}{2}$

$\therefore$ $\frac{dy}{dx}=0$