what is the slope of the line tangent to the curve y + 2 = x²/2 - 2sin y at the point (2, 0)? (a) -2 (b) 0…

Answer:

D. $\frac{2}{3}$

Explanation:

Step1: Differentiate both sides implicitly

Differentiate $y + 2=\frac{x^{2}}{2}-2\sin y$ with respect to $x$. The derivative of $y$ with respect to $x$ is $y'$, derivative of $2$ is $0$, derivative of $\frac{x^{2}}{2}$ is $x$ and derivative of $- 2\sin y$ is $-2\cos y\cdot y'$. So we get $y'=x - 2\cos y\cdot y'$.

Step2: Solve for $y'$

Rearrange the equation $y'=x - 2\cos y\cdot y'$ to isolate $y'$. Add $2\cos y\cdot y'$ to both sides: $y'+2\cos y\cdot y'=x$. Factor out $y'$: $y'(1 + 2\cos y)=x$. Then $y'=\frac{x}{1 + 2\cos y}$.

Step3: Substitute the point $(2,0)$

Substitute $x = 2$ and $y = 0$ into $y'=\frac{x}{1 + 2\cos y}$. Since $\cos(0)=1$, we have $y'=\frac{2}{1+2\times1}=\frac{2}{3}$.