Question

1. (5 pts.) find the derivative of the function $h(z)=sin(\frac{pi}{z})$, and compute $h(2)$.

Explanation:

Step1: Apply chain - rule

Let $u=\frac{\pi}{z}$, then $h(z)=\sin(u)$. The chain - rule states that $h^\prime(z)=\frac{d}{dz}(\sin(u))\cdot\frac{du}{dz}$. The derivative of $\sin(u)$ with respect to $u$ is $\cos(u)$, and $\frac{du}{dz}=\frac{d}{dz}(\frac{\pi}{z})=-\frac{\pi}{z^{2}}$. So $h^\prime(z)=\cos(\frac{\pi}{z})\cdot(-\frac{\pi}{z^{2}})=-\frac{\pi}{z^{2}}\cos(\frac{\pi}{z})$.

Step2: Evaluate $h^\prime(2)$

Substitute $z = 2$ into $h^\prime(z)$. We have $h^\prime(2)=-\frac{\pi}{2^{2}}\cos(\frac{\pi}{2})$. Since $\cos(\frac{\pi}{2}) = 0$, then $h^\prime(2)=-\frac{\pi}{4}\times0 = 0$.

Answer:

$h^\prime(z)=-\frac{\pi}{z^{2}}\cos(\frac{\pi}{z})$, $h^\prime(2)=0$