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i need help finding this with quotient rule (3x^4-3x)/(3x^4+3x)
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i know the formula just not to good at simplifiying
the fraction can be factorised and then the common term eliminated \[\frac{3x^4 - 3x}{3x^4 + 3x} = \frac{3x(x^3 -1)}{3x(x^3 +1)} = \frac{x^3 -1}{x^3 +1}\]
then u = x^3 - 1 du/dx = 3x^2 v = 3x+1 dv/dx = 3x^2
im sorry they want us to use quotient rulle for this
to find derivative
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yes so you are now using the quotient rule with a simplified algebraic fraction \[\frac{dy}{dx} =\frac{ (x^3 +1)\times3x^2 - (x^3 - 1)\times 3x^2}{(x^3 +1)^2}\] this simplifies to \[\frac{dy}{dx} = \frac{6x^2}{(x^3+1)^2}\]
do the simplifying at the start reduced the amount of simplification at the end...
oh thanks so much
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