Question #a1919

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Answer 1 · 2017-07-05T17:00:15

$F'(3)=152.$

Given that, $F(x)=5x^3+4x^2-7x+4,$ and we have to find $F'(3).$

We know that, $(x^n)'=nx^(n-1)...................(ast).$

Therefore, we have,

$F'(x)={5x^3+4x^2-7x+4}',$

$=(5x^3)'+(4x^2)'-(7x)'+(4)',$

$=5(x^3)'+4(x^2)'-7(x^1)'+0,$

$=5(3x^(3-1))+4(2x^(2-1))-7(1x^(1-1))............[because, (ast)],$

$=5(3x^2)+4(2x^1)-7(x^0),$

$:. F'(x)=15x^2+8x-7.$

$:. F'(3)=15(3)^2+8(3)-7=135+24-7,$

$rArr F'(3)=152.$