Prove, using the definition of a derivative, that if f(x) = cos x, then f'(x) = -sin x?

Question

I'd like to see work to get me through this problem, if possible, the actual steps of proving it. Any help welcome, thanks!

Answer 1

Generally, it's traditional to derive the derivative of sine first, and then to use the chain rule and the identities sin x = cos (π/2 -x) and cos x = sin (π/2 - x) to show that d/dx (cos x) = d/dx (sin (π/2 - x)) = cos (π/2 - x) * (-1) = -sin x. But if you want to go through the definition a second time:

d/dx (cos x)
[h→0]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)/h - [h→0]lim (sin x sin h)/h
cos x [h→0]lim (cos h - 1)/h - sin x [h→0]lim sin h/h
cos x [h→0]lim (cos h - 1)/h - sin x
cos x [h→0]lim (cos² h - 1)/(h(cos h + 1)) - sin x
cos x [h→0]lim -sin² h/(h(cos h + 1)) - sin x
cos x [h→0]lim -sin h/h * [h→0]lim sin h/(cos h + 1) - sin x
cos x * (-1) * 0 - sin x
-sin x

Note that this presumes you know [h→0]lim sin h/h = 1, which can be proved by using geometric arguments to establish the inequality cos h < sin h/h < 1 on the interval [-π/2, π/2]\{0}, and then invoking the squeeze theorem.

Answer 2

you can prove it using the definition of the derivative + the limits limx->0 sinx/x = 1 and limx->0 cosx-1/x =0 these are found in your text book assume f(x) = cosx then f '(x)= limh->0 cos(x+h) - cos(x) / h using trigonometric identities you substitute since cos (x+h) = cosxcosh-sinxsinh limh->0 cosxcosh-sinxsinh -cosx /h then you factor cosx out to get = limh->0 cosx(cosh-1) -sinxsinh /h = limh->0 cosx(cosh-1)/h - sinxsinh/h then you use the limits limx->0 sinx/x = 1 and limx->0 cosx-1/x =0 you will end up with limh->0 cosx(0) - sinx(1) = -sinx hope this helps good luck

Answer 3

lim...... (cos(x+h) - cos(x)) /h
h->0

expand the cosine

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

rearange:

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

split it up:

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

Now you should see two special limits:
lim h->0 (cos(h)-1)/h = 0
and
lim h->0 sin(h)/h = 1

so you are left with -sin(x)

Answer 4

f(x) = cos x

f(x+h) = cos(x+h)

= cosx cosh - sinx sinh

f(x+h) - f(x) = cosx cosh - sin x sin h - cosx

= cosx(cos h - 1) - sin x sin h

f(x+h)-f(x)/h = cosx[(cosh-1/h)] - sinx (sinh/h)

when h ->0 (cosh - 1)h = 0 and

sinh/h = 1

f'(x) = 0 - sinx

= -sin x

Answer 5

[cos(x+h) - cos(x)] / h

= [cos(x) cos(h) - sin(x) sin(h) - cos(x)] / h

= [cos(x) (cos(h) - 1)] / h - [sin(x) sin(h)] / h

Take limit as x -> 0

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

= -sin(x)