Answer:
The slope of the line tangent to the given curve at the point (2,0) is 2/3.
Step-by-step explanation:
The given equation is
\(y+2=(\frac{x^2}{2})-2\sin y\)
we need to find the slope of the line tangent to the given curve at the point (2,0).
\(slope=\frac{dy}{dx}_{(2,0)}\)
Differentiate given equation with respect to x.
\(\frac{dy}{dx}+0=\frac{2x}{2}-2(\cos y)\frac{dy}{dx}\)
\(\frac{dy}{dx}=x-2\cos y\frac{dy}{dx}\)
\(\frac{dy}{dx}+2\cos y\frac{dy}{dx}=x\)
\((1+2\cos y)\frac{dy}{dx}=x\)
Divide both sides by(1+2cos y).
\(\frac{dy}{dx}=\frac{x}{(1+2\cos y)}\)
Substitute x=2 and y=0.
\(\frac{dy}{dx}_{(2,0)}=\frac{2}{1+2\cos (0)}\)
\(\frac{dy}{dx}_{(2,0)}=\frac{2}{1+2(1)}\)
\(\frac{dy}{dx}_{(2,0)}=\frac{2}{3}\)
Therefore the slope of the line tangent to the given curve at the point (2,0) is 2/3.
