How do you differentiate #f(x)=(x^2+4x)*(2+x^-2)# using the product rule?

2 Answers
Feb 18, 2016

Answer:

#f'(x) = 4x + 8 - 4/x^2#

Explanation:

#f'(x) = frac{"d"}{"d"x}((x^2+4x)*(2+x^{-2}))#

#= (x^2+4x)*frac{"d"}{"d"x}(2+x^{-2}) #

#+ (2+x^{-2})*frac{"d"}{"d"x}(x^2+4x)#

#= (x^2+4x)*((-2)x^{-3}) #

#+ (2+x^{-2})*(2x+4)#

#= (-2/x-8/x^2) + (4x +2/x + 8 + 4/x^2)#

#= 4x + 8 - 4/x^2#

Feb 18, 2016

Answer:

#-2x^-2(x+4) + 2(2+x^-2)(x+2)#

Explanation:

using the #color(blue)(" product rule ")#

If f(x) = g(x).h(x) then f'(x) = g(x)h'(x) + h(x)g'(x)

hence #f'(x) = (x^2+4x) d/dx(2+x^-2) + (2+x^-2) d/dx(x^2+4x)#

#= (x^2+4x)(-2x^-3) + (2+x^-2)(2x+4)#

#= x(x+4)(-2x^-3) + (2+x^-2) 2(x+2) #
(removed 'common factors' from the 2 brackets of x and 2)

#rArr = -2x^-2(x+4) + 2(2+x^-2)(x+2)#