Jika h(x)=x²cotx, maka h’45° sama dengan..?mohon bantuannya.. :)

Turunan trigonometri
h'(45) –> h'(π/4)

h(x)= x² cot x
h '(x)= 2x (cot x) + x² (- cosec x)
h' (π/4) = 2(π/4) . cos π/4   -  (π/4)²  (cosec π/4)
h' (π/4) = π/2. (1/2 √2) – π²/16 . (√2)
h ' (π/4) = 1/4 π √2 – 1/16 π² √2
h'(π/4)=  π√2 ( 1/4 – 1/16 π)
h '(π/4) = π√2 (3/16  π)