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How do you differentiate #f(x) = 1/(x^3-4x)# using the quotient rule?

Answer 1

Christopher Allers

Quotient rule states that for a function #y=f(x)/g(x)#, #(dy)/(dx)=(f'(x)g(x)-f(x)g'(x))/g(x)^2#.

To be honest, I'd better use chain rule, here, but let's do as you asked.

#(df(x))/(dx)=((0)(x^3-4x)-1(3x^2-4))/(x^3-4x)^2#

#(df(x))/(dx)=(cancel((0)(x^3-4x))-1(3x^2-4))/(x^3-4x)^2#

#(df(x))/(dx)=(-1(3x^2-4))/(x^3-4x)^2=(4-3x^2)/(x^3-4x)^2#

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Answer 2

Christopher Agresta

To differentiate ( f(x) = \frac{1}{x^3 - 4x} ) using the quotient rule:

Identify ( u ) and ( v ): ( u = 1 ) ( v = x^3 - 4x )
Apply the quotient rule: ( f'(x) = \frac{u'v - uv'}{v^2} )
Find the derivatives of ( u ) and ( v ): ( u' = 0 ) ( v' = 3x^2 - 4 )
Substitute into the quotient rule formula: ( f'(x) = \frac{(0)(x^3 - 4x) - (1)(3x^2 - 4)}{(x^3 - 4x)^2} )
Simplify: ( f'(x) = \frac{-3x^2 + 4}{(x^3 - 4x)^2} )

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Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.