Proof of the Derivative of cos(x)

We use the definition of the derivative to prove the formula for the derivative of \( \cos(x) \). The derivative of the composite function \( \cos(u(x)) \) is also presented using the chain rule, with worked examples.

Proof Using the Definition of the Derivative

The definition of the derivative of a function \( f \) is

\[ f'(x)=\lim_{h\to0}\frac{f(x+h)-f(x)}{h} \]

Let \( f(x)=\cos x \). Then

\[ f'(x)=\lim_{h\to0}\frac{\cos(x+h)-\cos x}{h} \]

Using the identity

\[ \cos(x+h)=\cos x\cos h-\sin x\sin h \]

we obtain

\[ f'(x)=\lim_{h\to0}\frac{\cos x(\cos h-1)-\sin x\sin h}{h} \]

Separate the limits:

\[ f'(x)=\cos x\lim_{h\to0}\frac{\cos h-1}{h}-\sin x\lim_{h\to0}\frac{\sin h}{h} \]

Using the standard results:

\[ \lim_{h\to0}\frac{\sin h}{h}=1 \] \[ \lim_{h\to0}\frac{\cos h-1}{h}=0 \]

we get

\[ f'(x)=\cos x(0)-\sin x(1)=-\sin x \]

\[ \boxed{\dfrac{d}{dx}\cos x=-\sin x} \]

Graph of cos(x) and Its Derivative

The graphs of \( \cos x \) and its derivative are shown below. Maxima and minima of \( \cos x \) correspond to zeros of its derivative.

Graph of cos x and its derivative

Derivative of the Composite Function cos(u(x))

Using the chain rule:

\[ \frac{d}{dx}\cos(u(x))=\frac{d}{du}(\cos u)\cdot\frac{du}{dx} \] \[ =-\sin u\cdot\frac{du}{dx} \]

\[ \boxed{\frac{d}{dx}\cos(u(x))=-\sin(u(x))\,u'(x)} \]

Example

Find the derivatives:

  1. \( f(x)=\cos(2x+2) \)
  2. \( g(x)=\cos(\tan x) \)
  3. \( h(x)=\cos\!\left(\frac{x^2}{x^2+1}\right) \)

Solutions

  1. \[ u=2x+2,\quad u'=2 \] \[ f'(x)=-2\sin(2x+2) \]
  2. \[ u=\tan x,\quad u'=\sec^2 x \] \[ g'(x)=-\sin(\tan x)\sec^2 x \]
  3. \[ u=\frac{x^2}{x^2+1},\quad u'=\frac{2x}{(x^2+1)^2} \] \[ h'(x)=- \frac{2x}{(x^2+1)^2} \sin\!\left(\frac{x^2}{x^2+1}\right) \]

More References

Definition of the derivative
Squeezing theorem
Limits of trigonometric functions
Chain Rule