# How do you differentiate #f(x)=cotx(cscx)#?

This function can be differentiated using the "product rule":

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To differentiate ( f(x) = \cot(x) \cdot \csc(x) ), you can use the product rule of differentiation. The product rule states that if you have two functions ( u(x) ) and ( v(x) ), then the derivative of their product ( u(x) \cdot v(x) ) with respect to ( x ) is given by:

[ \frac{d}{dx} [u(x) \cdot v(x)] = u'(x) \cdot v(x) + u(x) \cdot v'(x) ]

Applying the product rule to ( f(x) = \cot(x) \cdot \csc(x) ), we have:

[ u(x) = \cot(x) ] [ v(x) = \csc(x) ]

[ u'(x) = -\csc^2(x) ] [ v'(x) = -\cot(x) \csc(x) ]

Substitute these into the product rule formula:

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

Simplify:

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

This is the derivative of ( f(x) ) with respect to ( x ).

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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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