How do you differentiate #(x-4)/(x^2+2)#?

2 Answers
Anthony R.
Dec 21, 2017

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

Explanation:

Apply the quotient rule which states:

#f(x)=g(x)/(h(x))->f'(x)=(g'(x)h(x)-h'(x)g(x))/((g(x))^2)#

Let

#g(x)=x-4#

#h(x)=x^2+2#

Thus,

#g'(x)=1#

#h'(x)=2x#

Now plugging into the formula:

#f'(x)=((1)*(x^2+2)-(2x)(x-4))/(x-4)^2#

Simplify:

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

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

Jim G.
Dec 21, 2017

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

Explanation:

#"differentiate using the "color(blue)"quotient rule"#

#"given "y=(g(x))/(h(x))" then"#

#dy/dx=(h(x)g'(x)-g(x)h'(x))/(h(x))^2larrcolor(blue)"quotient rule"#

#g(x)=x-4rArrg'(x)=1#

#h(x)=x^2+2rArrh'(x)=2x#

#rArrd/dx((x-4)/(x^2+2))#

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

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

Related questions
See all questions in Power Rule
Impact of this question
1410 views around the world
You can reuse this answer
Creative Commons License