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

Answer 1

Applying the derivative definition:

#dy/dx = lim_(h->0) (f(x+h)-f(x))/h#, where #h = deltax#

In our function, we substitute to obtain:

#lim_(h->0) (cos(x+h)-cos(x))/h#

Applying Trig's identity:

#cos(a+b) = cosacosb - sinasinb#,

we get:

#lim_(h->0) ((cosxcos h - sinxsin h)-cosx)/h#
Factoring out the #cosx# term, we get:
#lim_(h->0) (cosx(cos h-1) - sinxsin h)/h#

Two fractions can be separated from this:

#lim_(h->0) (cosx(cos h-1))/h - (sinxsin h)/h#

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

Here, the two that will be helpful are:

#lim_(x->0) sinx/x = 1#, and #lim_(x->0) (cosx-1)/x = 0#
Since those identities rely on the variable inside the functions being the same as the one used in the #lim# portion, we can only use these identities on terms using #h#, since that's what our #lim# uses. To work these into our equation, we first need to split our function up a bit more:
#lim_(h->0) (cosx(cos h-1))/h - (sinxsin h)/h#

becomes:

#lim_(h->0)cosx((cos h-1)/h) - sinx((sin h)/h)#

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

#lim_(h->0) cosx(0) - sinx(1)#

which is equivalent to:

#lim_(h->0) (-sinx)#
Since there are no more #h# variables, we can just drop the #lim_(h->0)#, giving us a final answer of: #-sinx#.
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Answer 2

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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Answer from HIX Tutor

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.

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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