# How do you differentiate #f(x)=cos5x * cot3x# using the product rule?

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To differentiate the function ( f(x) = \cos(5x) \cdot \cot(3x) ) using the product rule, we first identify the two functions being multiplied: ( \cos(5x) ) and ( \cot(3x) ). Then, we apply the product rule, which states that the derivative of the product of two functions is the derivative of the first function times the second function, plus the first function times the derivative of the second function.

Let ( u = \cos(5x) ) and ( v = \cot(3x) ).

Now, we find the derivatives of ( u ) and ( v ):

- ( \frac{du}{dx} = -5\sin(5x) )
- ( \frac{dv}{dx} = -3\csc^2(3x) )

Using the product rule, we have: [ \frac{d}{dx}[u \cdot v] = u \cdot \frac{dv}{dx} + \frac{du}{dx} \cdot v ] [ = \cos(5x) \cdot (-3\csc^2(3x)) + (-5\sin(5x)) \cdot \cot(3x) ] [ = -3\cos(5x)\csc^2(3x) - 5\sin(5x)\cot(3x) ]

Therefore, the derivative of ( f(x) = \cos(5x) \cdot \cot(3x) ) with respect to ( x ) using the product rule is ( -3\cos(5x)\csc^2(3x) - 5\sin(5x)\cot(3x) ).

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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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