Math Calculus
posted by Marissa .
Let f be a function such that
f(1)=3 and f'(1)=6
let h(x)=x^5f(x)
evaluate h'(x) at x=1

h'(x)=5*x^4*f(x)+x^5*f'(x)
h'(1)=5*(1)^4*f(1)+(1)^5*f'(1)=
5*1*3+(1)*6=156=9
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