# How do you differentiate #f(x)=sinx-4cos(5x)#?

The given function:

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So

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We essentially have the following:

We know:

We can plug this into the Chain Rule to get

We now have the following:

Hope this helps!

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To differentiate the function ( f(x) = \sin(x) - 4\cos(5x) ), you would apply the differentiation rules for trigonometric functions and chain rule:

- Differentiate ( \sin(x) ) to get ( \cos(x) ).
- Differentiate ( -4\cos(5x) ) using the chain rule: [ \frac{d}{dx} \left( -4\cos(5x) \right) = -4 \times \frac{d}{dx} \left( \cos(5x) \right) = -4 \times (-5\sin(5x)) = 20\sin(5x) ]

Therefore, the derivative of ( f(x) ) is ( f'(x) = \cos(x) - 20\sin(5x) ).

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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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