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.#