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

Answer 1

Christopher Allers

The derivative is #=((xe^x-1-2x^3-e^x))/(x^3-e^x-1)^2#

The rule of the quotient is

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

We have

#f(x)=(x)/(x^3-e^x-1)#

Here,

#u(x)=x#, #=># #u'(x)=1#

#v(x)=x^3-e^x-1#, #=>#, #v'(x)=3x^2-e^x#

Consequently,

#f'(x)=(1*(x^3-e^x-1)-x(3x^2-e^x))/(x^3-e^x-1)^2#

#=((x^3-e^x-1-3x^3+xe^x))/(x^3-e^x-1)^2#

#=((xe^x-1-2x^3-e^x))/(x^3-e^x-1)^2#

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Answer 2

Ezra Adams

To differentiate ( f(x) = \frac{x}{x^3 - e^x - 1} ) using the quotient rule, you would apply the formula ( \left(\frac{u}{v}\right)' = \frac{u'v - uv'}{v^2} ), where ( u = x ) and ( v = x^3 - e^x - 1 ). Then, differentiate ( u ) and ( v ) separately, and substitute into the formula.

( u' = 1 ) and ( v' = 3x^2 - e^x )

Now apply the quotient rule:

( f'(x) = \frac{(1)(x^3 - e^x - 1) - (x)(3x^2 - e^x)}{(x^3 - e^x - 1)^2} )

Finally, simplify the expression.

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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.