# How do you differentiate # g(x) = 2 csc ^2x + 5 cos x #?

It helps to know some trig derivatives:

We're also going to need product rule since there's a

Now take the derivative with product rule:

Simplify:

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To differentiate ( g(x) = 2 \csc^2(x) + 5 \cos(x) ), use the following steps:

- Recall that the derivative of (\csc(x)) is (-\csc(x) \cot(x)).
- The derivative of (\cos(x)) is (-\sin(x)).

Now, apply these derivatives to ( g(x) = 2 \csc^2(x) + 5 \cos(x) ):

( g'(x) = 2 \left( -\csc(x) \cot(x) \right) - 5 \sin(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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