the proof of the derivative of cos x = - sin x?

Question

where can I find the proof of the derivative of cos x = - sin x

Answer 1

It depends on what knowledge you're given.
Since, in most Calculus books, the derivative of sin(x) is first proven prior to the derivative of cos(x), let's assume we know the derivative of sin(x) as cos(x).

Then

d/dx cos(x) =
d/dx sin(pi/2 - x) =

Using the chain rule,

cos(pi/2 - x) (-1)
-cos(pi/2 - x)
- [cos(pi/2)cos(x) + sin(pi/2)sin(x)]
- [0*cos(x) + (1)sin(x)]
- [0 + sin(x)]
-sin(x)

-OR-

Using the definition of the derivative,

lim [ cos(x + h) - cos(x) ] / h
h -> 0

lim [ cos(x)cos(h) - sin(x)sin(h) - cos(x) ] / h
h -> 0

lim [ cos(x)cos(h) - cos(x) - sin(x)sin(h) ] / h
h -> 0

lim [ cos(x) [ cos(h) - 1 ] - sin(x)sin(h) ] / h
h -> 0

lim [ cos(x) [ cos(h) - 1 ]/h - [sin(x)sin(h)]/h
h -> 0

lim [ cos(x) [cos(h) - 1]/h ] - lim sin(x)sin(h)/h
h -> 0 . . . . . . . . . . . . . . . . . . h -> 0

Factor cos(x) from the first limit, and sin(x) from the second limit.

cos(x) lim [ cos(h) - 1 ]/h - sin(x) lim sin(h)/h
. . . . . . h -> 0 . . . . . . . . . . . . . . . . . h -> 0

We need foreknowledge of trig limits to realize that
lim [ cos(h) - 1 ]/h = 0, and
h -> 0
lim sin(h)/h = 1
h -> 0

The above then becomes

cos(x) (0) - sin(x)(1)
0 - sin(x)

-sin(x)

Answer 2

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the proof of the derivative of cos x = - sin x?
where can I find the proof of the derivative of cos x = - sin x

Answer 3

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cos(x+h) - cos(x) = cos(x)(cos(h) - 1) - sin(x)sin(h) That is, if we show it this way, then we need to prove that (1) lim(h->0) ( cos(h) - 1 ) / h = 0 (2) lim(h->0) sin(h) / h = 1 Both of these are true, but neither is particularly easy to prove without the standard "use L'hopital's rule" fallacious step. If you know anything else, like the fact that the derivative of sin(x) is cos(x), then there are easier ways; however, I don't want to write a lot here without knowing what you can use.

Answer 4

The easiest way to do this is to write down cos x as a power series, and differentiate each term one by one.

After the first few terms you will see a pattern emerge, and this pattern is -sin x.

Answer 5

In addition to the previous correct answers, I will add:

sinx(x) = [exp(ix) - exp(-ix)]/(2i)
cos(x) = [exp(ix) + exp(-ix)]/2

dexp(ax)/dx = aexp(ax)

dcos(x)/dx = [iexp(ix) - iexp(-ix)]/2
=[-exp(ix) + exp(-ix)]/(2i) = -sin(x)

And another approach that isn't as straightforward....

cos(2x) = cos^2(x) - sin^2(x)
1 = cos^2(x) + sin^2(x)

1 + cos(2x) = 2cos^2(x)

differentiate both sides and call the derivative of cos(x) the function g(x).

2g(2x) = 4cos(x)g(x)
g(2x)/g(x) = 2cos(x)

Now note: sin(2x) = 2sin(x)cos(x)
or sin(2x)/sin(x) = 2cos(x)

So g(x) = sin(x) or -sin(x) (sign lost in the g(2x)/g(x) ratio)

Noting that cos(0) = 1, is a local maximum which thus decreases as x increases from zero, while sin(x) increases and is positive, we know that g(x) = -sin(x) and not sin(x).

The other answers are more straightforward and conventional but I didn't want to be a repetition of them.

Answer 6

f `(x) = lim h->0 [ cos (x + h) - cos x ] / h
= lim h->0 [ (- 2.sin[2x + h) / 2)] .sin (h/2) ] / h
= - lim h->0 [sin(h/2) /(h/2)].(sin ((2x + h) / 2)]
= - 1.sin x
= - sin x

Answer 7

Assuming that limit h -> 0
y=cos x.
dy/dx
= (cos(x+h) - cosx)/h
=(cos x*cos h - sin x*sin h -cos x)/h
As h is tending to zero..... cos h =1
=(cos x - cos x -sin x*sin h)/h
= -sin x*(sin h)/h
sin h/h=1
= - sin x

Answer 8

sorry tried but couldnt get the answer