# How do you find the derivative of #y=cos(x)# from first principle?

Applying the derivative definition:

In our function, we substitute to obtain:

Applying Trig's identity:

we get:

Two fractions can be separated from this:

The harder part is about to come: identifying known formulas.

Here, the two that will be helpful are:

becomes:

With the help of the previously established formulas, we now have:

which is equivalent to:

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To find the derivative of ( y = \cos(x) ) from first principles, we start with the definition of the derivative:

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

For ( y = \cos(x) ), we substitute into this formula:

[ \begin{split} f'(x) & = \lim_{h \to 0} \frac{\cos(x + h) - \cos(x)}{h} \ & = \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(\frac{h}{2})}{h} \ & = \lim_{h \to 0} \frac{-2\sin^2(\frac{h}{2})}{\frac{h}{2}} \cdot \frac{\frac{1}{2}}{\frac{1}{2}} \ & = \lim_{h \to 0} \frac{-2\sin^2(\frac{h}{2})}{\frac{h}{2}} \cdot \lim_{h \to 0} \frac{1}{2} \ & = -\lim_{h \to 0} \frac{2\sin^2(\frac{h}{2})}{\frac{h}{2}} \ & = -\lim_{h \to 0} 4\frac{\sin(\frac{h}{2})}{\frac{h}{2}} \frac{\sin(\frac{h}{2})}{\frac{h}{2}} \ & = -4 \cdot 1 \cdot 1 \ & = -4 \end{split} ]

So, ( \frac{d}{dx}(\cos(x)) = -\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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