# How do you use the definition of a derivative to find the derivative of #f(x)=cosx#?

By definition:

We can now use the trigonometric identity:

and obtain:

that is:

But:

and:

and we can conclude:

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To find the derivative of ( f(x) = \cos(x) ) using the definition of a derivative, you apply the limit definition of the derivative, which states:

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

For ( f(x) = \cos(x) ), substitute ( \cos(x) ) into the definition:

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

Using the trigonometric identity ( \cos(a + b) = \cos(a)\cos(b) - \sin(a)\sin(b) ), we rewrite the numerator:

[ 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} ]

[ = \lim_{h \to 0} \frac{-2\sin^2(h/2)\sin(x)}{h} ]

Now, using the trigonometric limit ( \lim_{\theta \to 0} \frac{\sin(\theta)}{\theta} = 1 ), we can evaluate the limit:

[ f'(x) = -\sin(x) \lim_{h \to 0} \frac{2\sin(h/2)}{h/2} ]

[ = -\sin(x) \cdot 2 ]

[ = -2\sin(x) ]

So, the derivative of ( f(x) = \cos(x) ) is ( f'(x) = -2\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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