Question

1. (5 pts.) find the derivative of the function $h(z)=sin(\frac{1}{z})$, and compute $h(2)$. 3. (5 pts.) consider the curve described by the equation $2y^{2}-x^{2}=4y$. verify that the point $(4, - 2)$ lies on this curve, and find $y$ at that point.

Explanation:

2.

Step1: Apply chain - rule

Let $u=\frac{1}{z}=z^{-1}$. Then $h(z)=\sin(u)$. The chain - rule states that $h^\prime(z)=\frac{dh}{du}\cdot\frac{du}{dz}$. First, $\frac{dh}{du}=\cos(u)$ and $\frac{du}{dz}=-z^{-2}=-\frac{1}{z^{2}}$.

Step2: Substitute $u$ back

Since $u = \frac{1}{z}$, we have $h^\prime(z)=\cos(\frac{1}{z})\cdot(-\frac{1}{z^{2}})=-\frac{1}{z^{2}}\cos(\frac{1}{z})$.

Step3: Compute $h^\prime(2)$

Substitute $z = 2$ into $h^\prime(z)$. $h^\prime(2)=-\frac{1}{2^{2}}\cos(\frac{1}{2})=-\frac{1}{4}\cos(\frac{1}{2})$.

Step1: Verify the point lies on the curve

Substitute $x = 4$ and $y=-2$ into the equation $2y^{2}-x^{2}=4y$.
Left - hand side: $2(-2)^{2}-4^{2}=2\times4 - 16=8 - 16=-8$.
Right - hand side: $4\times(-2)=-8$. Since the left - hand side equals the right - hand side, the point $(4,-2)$ lies on the curve.

Step2: Differentiate implicitly

Differentiate the equation $2y^{2}-x^{2}=4y$ with respect to $x$.
Using the chain - rule for the $y$ terms, we get $4y\cdot y^\prime-2x = 4y^\prime$.

Step3: Solve for $y^\prime$

Rearrange the terms: $4y\cdot y^\prime-4y^\prime=2x$. Factor out $y^\prime$: $y^\prime(4y - 4)=2x$. Then $y^\prime=\frac{2x}{4y - 4}=\frac{x}{2y - 2}$.

Step4: Evaluate $y^\prime$ at the point $(4,-2)$

Substitute $x = 4$ and $y=-2$ into $y^\prime$. $y^\prime=\frac{4}{2\times(-2)-2}=\frac{4}{-4 - 2}=-\frac{2}{3}$.

Answer:

$h^\prime(z)=-\frac{1}{z^{2}}\cos(\frac{1}{z})$, $h^\prime(2)=-\frac{1}{4}\cos(\frac{1}{2})$

3.