Differentiate:  f(x) = ((x-7)(x^2+3X)) / x^3
in Calculus Answers by Level 1 User (180 points)
This answer does not coincide with the answers I was given to choose from:
f'(x) = x - (42/x^2) - (42/x^3)
f'(x) = 42 + (42/x)
f'(x) = (10/x^2) - (42/x^3)
f'(x) = (4/x^2) + (42/x^3)

I am really confused.

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1 Answer

Differentiate: f(x) = ((x-7)(x^2+3X)) / x^3

you do this in 2 steps.  first differentiate a product (in the paranthesis).  then differentiate a difference.

product f(x) = u(x)*v(x), difference: f(x) = u(x)/v(x)

f'(x) = u(x)*v'(x) + v(x)*u'(x)           f'(x) = [v(x)*u'(x) - u(x)*v'(x)]/[v(x)^2]

step 1: u(x) = x-7 v(x) = x^2 + 3x

            u'(x) = 1   v'(x) = 2x + 3

plug into formula above

f'(x) = (x-7)(2x + 3) + (x^2 + 3x)(1)

f'(x) = 2x^2 + 3x -14x -21 + x^2 + 3x

f'(x) = 3x^2 - 8x - 21

step 2: u(x) = 3x^2 - 8x - 21 v(x) = x^3

            u'(x)= 6x - 8              v'(x)= 3x^2

f'(x) = {[x^3 (6x - 8)] - [(3x^2 - 8x - 21)(3x^2)]}/(x^3)^2

f'(x) = {6x^4 -8x^3 - [9x^4 -24x^3 - 63x^2]}/x^6

f'(x) = -3x^4 + 16x^3 - 63x^2/x^6

f'(x) = x^2(-3x^2 + 16x -63)/x^6

f'(x) = -3x^2 + 16x - 63/x^4

 

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