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March 29, 2017

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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

  • Math- Calculus - ,

    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=15-6=9

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