# Find the derivative of #cos# using First Principles?

Using the cosine sum of angle formula:

We get

We now have to rely on some standard calculus limits:

And so using these we have:

Hence,

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To find the derivative of ( \cos(x) ) using first principles:

- Start with the limit definition of the derivative:

[ f'(x) = \lim_{h \to 0} \frac{f(x + h) - f(x)}{h} ]

- Substitute ( f(x) = \cos(x) ) into the limit definition:

[ f'(x) = \lim_{h \to 0} \frac{\cos(x + h) - \cos(x)}{h} ]

- Use the trigonometric identity ( \cos(a + b) = \cos(a)\cos(b) - \sin(a)\sin(b) ):

[ f'(x) = \lim_{h \to 0} \frac{(\cos(x)\cos(h) - \sin(x)\sin(h)) - \cos(x)}{h} ]

- Expand and simplify the expression:

[ f'(x) = \lim_{h \to 0} \frac{\cos(x)\cos(h) - \sin(x)\sin(h) - \cos(x)}{h} ] [ = \lim_{h \to 0} \frac{\cos(x)(\cos(h) - 1) - \sin(x)\sin(h)}{h} ]

- Apply the limit laws and trigonometric limit properties to evaluate the limit:

[ f'(x) = -\sin(x) \lim_{h \to 0} \frac{\sin(h)}{h} + \cos(x) \lim_{h \to 0} \frac{\cos(h) - 1}{h} ]

- Recognize that ( \lim_{h \to 0} \frac{\sin(h)}{h} = 1 ) and ( \lim_{h \to 0} \frac{\cos(h) - 1}{h} = 0 ):

[ f'(x) = -\sin(x) \cdot 1 + \cos(x) \cdot 0 ] [ = -\sin(x) ]

Therefore, the derivative of ( \cos(x) ) with respect to ( x ) is ( -\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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