How do you find the derivative of #5/((x^2) + x + 1)^2#?

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Answer 1 · 2018-03-12T12:51:23

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

For a question like this, you must use the quotient rule...

If #f(x) = color(red)(u) /color(green)(v)# then...

#f'(x) = (color(green)(v) color(red)(u') - color(green)(v') color(red)(u))/ color(green)(v^2)#

So for the given function

#f(x)= color(red)(5)/color(green)((x^2 +x +1)^2)#

#f'(x) = (color(green)((x^2 +x +1)^2) color(red)(0) - color(green)(2(x^2+x+1)) color(red)(5))/ color(green)((x^2 +x +1)^2)#

#=> f'(x) = (-10(x^2+x+1))/((x^2 +x +1)^2)#

#=> f'(x) = (-10)/((x^2+x+1)^2)#