How do you show derivative of sin x equal to cos x by definition?

Answer 1

sin(x + dx) = sinxcosdx + sindxcosx

d(sinx) / dx = ( (sinxcosdx + sindxcosx) - sinx ) / dx

As dx very small
sindx = dx
cosdx = 1

Therefore:

d(sinx) / dx = ( (sinx(1) + dx cosx) - sinx ) / dx

= (sinx - sinx + dx cosx) / dx

= dx cosx / dx
= cosx

Answer 2

The answer primarily lies in limit laws. The two limits you need to know are:

1) lim (x -> 0, sin(x)/x) = 1, and
2) lim (x -> 0, (cos(x) - 1)/x = 0.

You would also need to know the sine addition law;
sin(a + b) = sin(a)cos(b) + sin(b)cos(a).

Let's start the proof.

Let f(x) = sin(x)

f'(x) = lim { f(x + h) - f(x) } / h
h -> 0

lim { sin(x + h) - sin(x) } / h
h -> 0

lim { [sinxcosh + sinhcosx] - [sinx] } / h
h -> 0

Rearranging the terms, i.e moving sinx between the two terms in the first set of brackets,

lim { [sinxcosh - sinx] + sinhcosx] } / h
h -> 0

Now, we can split this up into two separate limits and two fractions.

lim { [sinxcosh - sinx] } / h + lim (h -> 0, sinhcosx)/h
h -> 0

Due to text constraints, I'm going to work these limits out individually. I'll call the first limit A and the second limit B, and our answer should be A + B.

A = lim { [sinxcosh - sinx] } / h
h -> 0

Remember that h is our limit variable, and everything else is treated as a constant. This means we can effectively factor sin(x) out of the entire limit.

A = sin(x) * lim (h -> 0, [cosh - 1]/h)

But remember that we establishd that this limit is equal to 0. Therefore,

A = sin(x) * 0 = 0

B = lim (sinhcosx/h)
h -> 0

We can factor out cos(x) out of the limit because it's consider a constant (and there are NO h terms in there.

B = cos(x) * lim (h -> 0, sinh/h)

But remember, that we established right at the start that this limit is equal to 1. Therefore

B = cos(x) * 1 = cos(x)

The whole limit is equal to A + B, or
0 + cos(x) = cos(x).

Therefore, the derivative of sin(x) is cos(x).

As you can see, it was imperative that we used these two facts:

1) lim (x -> 0, sin(x)/x) = 1, and
2) lim (x -> 0, (cos(x) - 1)/x = 0

The actual proof of these two limits involves something called the Squeeze Theorem, and the only part of the proof I omitted was the details of why the above 2 limits are true.

Answer 3

An easy way is to look at the taylor series of sinx around 0 and take the derivative of this infinite polynomial and you will find the taylor series for cos x.

Answer 4

let y=sinx
y+delta y=sin(x+deltax)
where a small increment of delta x in xresults in a small increment of delta y in y
y+delta y-y=sin(x+delta x)-sinx
delta y=2cos(x+deltax/2)sin(delta x/2)
dividing both sides by delta x and allowing delta x to tend to zero
lim.delta x tending to zerodelta y/delta x
=limit delta x tending to zero 2cos(x+deltax/2)sin(deltax/2)/deltax
the LHS id dy/dx by definition
the RHS=lim delta x tending to zero cos(x+deltax/2)sindeltax/2/deltax/2
=cosx

Answer 5

There are a number of ways to do this.
One is to use the identity that
sinx = (e^ix - e^-ix)/2 and differentiate that.

Answer 6

The definition of derivate is:

df(x) = lím..... f(x+Δx) - f(x)
dx....... Δx→0........Δx

using that for the sin:

d (sinx) = lím..... sin(x+Δx) - sinx
dx .......... Δx→0...........Δx

sin(a+b) = sina*cosb + cosa*sinb
then:
sin(x+Δx) = sinx*cos(Δx) + cosx*sin(Δx)

by substitution:

d (sinx) = lím...... sinx*cos(Δx) + cosx*sin(Δx) - sinx
dx............Δx→0 ...................... Δx

agroupping:

d (sinx) = lím....... cosx*sin(Δx) - sinx (1 - cos(Δx))
dx............ Δx→0.................... Δx

arranging in fractions:

d (sinx) = lím ...... / cosx *. /sin(Δx)\ - sinx *. /.1 - cos(Δx) \ \
dx ............Δx→0.. \...........\ .. Δx ./..............\........ Δx .. / /

d (sinx) =. cosx * lím ... /sin(Δx)\ - sinx * lím... /.1 - cos(Δx).\
dx.....................Δx→0 \ ..Δx.. /............Δx→0\...... Δx ...... /

This límits have by definition this values:

. lím ... /sin(Δx)\ = 1
Δx→0 \... Δx.../

. lím ... /.1 - cos(Δx) \ = 0
Δx→0.. \...... Δx ..... /

then:

d (sinx) = cosx * 1 - sinx * 0
dx

finally

d (sinx) = cosx
dx