How do you find f'(x) using the definition of a derivative for #f(x)=cos x#?
See the explanation.
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The answer is
This is the definition of the derivative:
So we know that
Now, before we progress any further, we need to lay down two facts:

#lim_(x>h) (cos(h)1)/h = 0# 
#lim_(x>h) sin(h)/h = 1# If you'd like to verify these yourself, you can just plug these functions into your calculator, and see what they approach as
#x# nears 0.Now, if you noticed,
#cos(x+h)# is not very nice to work with as it is. So, we'll need to use an identity to split it up. If you look it up, you'll know that:#cos(x+h) = cos(x)cos(h)  sin(x)sin(h)# So, let's use this to split up
#cos(x+h)# :#lim_(x>h)(cos(x)cos(h)  sin(x)sin(h)  cos(x))/h# Now we can factor out a
#cos(x)# out of 2 of the terms, and we'll get:#lim_(x>h)(cos(x)(cos(h)1)  sin(x)sin(h))/h# And if we did some algebra:
#lim_(x>h)(cos(x)(cos(h)1))/h  (sin(x)sin(h))/h# And applied our first identity:
#lim_(x>h)cos(x)(0)  (sin(x)sin(h))/h# And now if we factored out a
#sin(x)# , we'd get:#lim_(x>h)  sin(x)(sin(h)/h)# Now we can apply our second identity:
#lim_(x>h)  sin(x)(1)# And since there are no
#h# terms in this, the limit just cancels out, and we're left with# sin(x)# .Of course, this is the ultralong way to derive the answer to this, and most calculus teachers just prefer you simply memorise this. And practically speaking, it's easier than deriving it all over each time.
The first 11 minutes of this video do this too. The only difference is that this video derives
#lim_(x>h)(sin(x+h)  sin(x))/h# , but to be honest the process is nearly completely the same.Hope that helps :)
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To find ( f'(x) ) using the definition of a derivative for ( f(x) = \cos(x) ), you apply the definition of the derivative:
[ f'(x) = \lim_{h \to 0} \frac{f(x + h)  f(x)}{h} ]
Substitute ( f(x) = \cos(x) ) into the definition:
[ f'(x) = \lim_{h \to 0} \frac{\cos(x + h)  \cos(x)}{h} ]
Then 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} ]
Factor out ( \cos(x) ) from the first term in the numerator:
[ f'(x) = \lim_{h \to 0} \frac{\cos(x)(\cos(h)  1)  \sin(x)\sin(h)}{h} ]
Apply the identity ( \lim_{h \to 0} \frac{\sin(h)}{h} = 1 ):
[ f'(x) = \sin(x) ]
Therefore, ( f'(x) = \sin(x) ).
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When evaluating a onesided 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 onesided 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 onesided 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 onesided 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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