Question 1149907
<pre>
Consider the function ℎ(𝑥) = 1 + sin(𝜋𝑥)
Find ℎ′(1).

ℎ(𝑥) =  1 + sin(𝜋𝑥)
ℎ'(𝑥) = 0 + 𝜋∙cos(𝜋𝑥)
ℎ'(1) = 𝜋∙cos(𝜋∙1)
ℎ'(1) = 𝜋∙cos(𝜋)
ℎ'(1) = 𝜋∙(-1)
ℎ'(1) = -𝜋

What's all that about g?  g is not involved with h or h'

Edwin</pre>