How do you find the derivative of #f(x)=x^3/(x^3-5x+10)#?

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Answer 1 · 2018-03-25T16:58:49

#=-10(x^2(x-3))/(x^3-5x+10)^2#

Let #u=f(x)# and #v=g(x)#

Then #d/dx(u/v)=(u'v-uv')/v^2#

For this question:

#u=x^3#
#u'=3x^2#

#v=x^3-5x+10#
#v'=3x^2-5#

#d/dx(u/v)=(3x^2(x^3-5x+10)-x^3(3x^2-5))/(x^3-5x+10)^2#

#=(cancel(3x^5)-15x^3+30x^2cancel(-3x^5)+5x^3)/(x^3-5x+10)^2#

#=(-10x^3+30x^2)/(x^3-5x+10)^2#

#=-10(x^2(x-3))/(x^3-5x+10)^2#