# How do you differentiate #f(x)=(x^3-x)(cotx-2x)# using the product rule?

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To differentiate ( f(x) = (x^3 - x)(\cot(x) - 2x) ) using the product rule, you apply the formula:

[ (uv)' = u'v + uv' ]

where ( u = x^3 - x ) and ( v = \cot(x) - 2x ).

( u' = 3x^2 - 1 ) (differentiating ( u ) with respect to ( x ))

( v' = -\csc^2(x) - 2 ) (differentiating ( v ) with respect to ( x ))

Thus, using the product rule:

[ f'(x) = (3x^2 - 1)(\cot(x) - 2x) + (x^3 - x)(-\csc^2(x) - 2) ]

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To differentiate ( f(x) = (x^3 - x)(\cot(x) - 2x) ) using the product rule, let's denote the two functions as ( u(x) = x^3 - x ) and ( v(x) = \cot(x) - 2x ). Then, applying the product rule ( (uv)' = u'v + uv' ), we find:

( u'(x) = 3x^2 - 1 ) and ( v'(x) = -\csc^2(x) - 2 ).

Now, we can use the product rule:

( f'(x) = u'v + uv' = (3x^2 - 1)(\cot(x) - 2x) + (x^3 - x)(-\csc^2(x) - 2) ).

So, the derivative of ( f(x) ) with respect to ( x ) is:

( f'(x) = (3x^2 - 1)(\cot(x) - 2x) + (x^3 - x)(-\csc^2(x) - 2) ).

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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.

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.

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.

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