# Why is the differential of cos x equal to -sin x?

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### 3 Answers

We'll take the function f(x) = cos x.

Now, we'll consider the ratio:

R(x) = [f(x)-f(x0)]/(x-x0)

We'll substitute f(x) and f(x0) by the expression of the function:

f(x) = cos x

f(x0) = cos x0.

[f(x)-f(x0)] = cos x - cos x0

We'll transform the difference into a product:

cos x - cos x0 = -2 sin [(x+x0)/2]*sin [(x-x0)/2]

If we'll calculate the limit of the ratio R(x), we'll get the value of the first derivative of the function, in the point x = x0.

lim R(x) = lim -2 sin [(x+x0)/2]*sin [(x-x0)/2]/(x-x0)

lim R(x) = -lim sin [(x+x0)/2]*lim sin [(x-x0)/2]/(x-x0)/2

But lim sin a/a = 1

lim sin [(x-x0)/2]/(x-x0)/2 = 1

lim R(x) = -lim sin [(x+x0)/2]*1

lim R(x) = -lim sin [(x+x0)/2]

-lim sin [(x+x0)/2] = -sin (x0+x0)/2 = -sin 2x0/2 = -sin x0

But lim R(x) = f'(x0)

**f'(x0) = -sin x0, when f(x) = cos x.**

The definition of the differential coefficient of any function f(x) is the limitting ratio the increment in f(x) and the x.

So d/dx f(x) = f'(x)= lt delta x-->0 {f(x+deltax0-f(x)}/delta x.

So by definition d/dx (cosx)' = Lt deltx-->0{cos(x+deltax) -cosx}/deltax

= Lt deltax -->0 {cosx*cosxdeltax-sin*sindeltax*indeltx) -cosx}/deltax.

= Lt deltax -->0 {(cosx)(cosdeltax - 1) - sinx*sin delta x}/deltax.

=Lt deltax --> 0 (cosx)(cosdeltax - 1)/deltax - L(sinx*{(sindetax)/deltax

= (cosx )(0) - (sinx)(1) , as by theorem ,lt a--> 0 (sina)/a = 1 and Lt a--> 0 (1-cosa)/a = 0.

= -sinx.

This establishes d/dx(cosx) = -sinx.

For a function f(x) the differential can be derived from the value we get for limit h-->0 [ f(x+h)- f(x)] / h.

Now cos( a+b)= cos a* cos b - sin a * sin b

For the function f(x) = cos x.

f(x+h)= cos (x+h) = cos x* cos h - sin x * sin h.

So limit h-->0 [ f(x+h)- f(x)] / h

= limit h--->0 [ cos x* cos h - sin x * sin h- cos x] / h

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

= cos x * 0 - sin x * 1

= -sin x.