# How do you find the derivative of #f(x) = xcot(x)#?

Use the product rule:

We get:

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To find the derivative of ( f(x) = x \cot(x) ), we 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 is given by ( \frac{d}{dx}[u(x)v(x)] = u'(x)v(x) + u(x)v'(x) ).

Applying the product rule to ( f(x) = x \cot(x) ), where ( u(x) = x ) and ( v(x) = \cot(x) ), we have:

[ f'(x) = u'(x)v(x) + u(x)v'(x) ]

Now, differentiate ( u(x) = x ) to get ( u'(x) = 1 ), and differentiate ( v(x) = \cot(x) ) using the chain rule, which states that if ( g(x) ) is differentiable at ( x ) and ( h(x) ) is differentiable at ( g(x) ), then ( (h \circ g)'(x) = h'(g(x)) \cdot g'(x) ). In this case, ( g(x) = x ) and ( h(x) = \cot(x) ), so ( h'(x) = -\csc^2(x) ) and ( g'(x) = 1 ).

Now, substitute these derivatives into the product rule formula:

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

Simplifying, we get:

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

So, the derivative of ( f(x) = x \cot(x) ) is ( f'(x) = \cot(x) - x\csc^2(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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