How do you differentiate #f(x)= (9-x)(5/(x^2 -4))# using the product rule?

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Answer 1 · 2018-05-26T07:51:38

#f'(x)=5*(x^2-18x+4)/((x-2)^2*(x+2)^2)#

After the product rule
#(uv)'=u'v+uv'#
we get

#f'(x)=-5/(x^2-4)+(9-x)*(-5)*(x^2-4)^(-2)*2*x#

which simplifies to
#f'(x)=5*(x^2-18x+4)/((x-2)^2*(x+2)^2)#

Answer 2 · 2018-05-26T08:06:22

#f'(x)=(5(x^2 -18x+4))/(x^2 -4)^2 #

#f(x)= (9-x)(5/(x^2 -4))#

#f'(x)= [d/dx(9-x)] (5/(x^2 -4))+[d/dx(5/(x^2 -4))] (9-x)#

Differentiate inner terms,

#f'(x)=(-1) (5/(x^2 -4))+(-(10x)/(x^2 -4)^2) (9-x)#

Simplify,

#f'(x)=-5/(x^2 -4)-(10x(9-x))/(x^2 -4)^2 #

Common denominator,

#f'(x)=-(5(x^2 -4))/(x^2 -4)^2-(10x(9-x))/(x^2 -4)^2 #

Simplify,

#f'(x)=-(5x^2 -20)/(x^2 -4)^2-(90x-10x^2)/(x^2 -4)^2 #

Combine,

#f'(x)=(-5x^2 +20-90x+10x^2)/(x^2 -4)^2 #

Simplify,

#f'(x)=(5x^2 -90x+20)/(x^2 -4)^2 #

Factorise,

#f'(x)=(5(x^2 -18x+4))/(x^2 -4)^2 #