The way you typed it, sinπ would just be a constant, and
h ' (x) = -sinπ/x^2
I will guess that you meant
h(x) = sin (π/x), which would be a totally different function
h '(x) = cos(π/x) ( -π/x^2)
so h ' (4) = (-π/16)cos(π/4)
I will let you do the button-pushing.
if h(x)=sinπ/x then find hderivative(4)
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