how the slope is possibly solved?
the slope of the line tangent to the curve x + y^2 = 1 at (0,1)
how does it look like?
anybody....?
Take the given equation:
$\displaystyle x+y^2=1$
Implicitly differentiate with respect to x:
$\displaystyle 1+2y\frac{dy}{dx}=0$
$\displaystyle \frac{dy}{dx}=-\frac{1}{2y}$
At $\displaystyle y=1$ we then have:
$\displaystyle \frac{dy}{dx}\left|_{y=1}=-\frac{1}{2}$
and this is the slope of the line tangent to the curve at the given point.