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, complete with worked examples.
Proof Using the Definition of the Derivative
The definition of the derivative of a function \( f \) is:
\[ f'(x) = \lim_{h\to 0} \frac{f(x+h) - f(x)}{h} \]Let \( f(x) = \cos x \). Then:
\[ f'(x) = \lim_{h\to 0} \frac{\cos(x+h) - \cos x}{h} \]Using the trigonometric identity:
\[ \cos(x+h) = \cos x \cos h - \sin x \sin h \]we obtain:
\[ f'(x) = \lim_{h\to 0} \frac{\cos x(\cos h - 1) - \sin x \sin h}{h} \]Separate the limits:
\[ f'(x) = \cos x \lim_{h\to 0} \frac{\cos h - 1}{h} - \sin x \lim_{h\to 0} \frac{\sin h}{h} \]Using the standard results:
\[ \lim_{h\to 0} \frac{\sin h}{h} = 1 \] \[ \lim_{h\to 0} \frac{\cos h - 1}{h} = 0 \]we get:
\[ f'(x) = \cos x(0) - \sin x(1) = -\sin x \]
Formula:
\[ \frac{d}{dx} \cos x = -\sin x \]
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.
-and-its-derivative.gif)
Derivative of the Composite Function cos(u(x))
Using the chain rule:
\[ \frac{d}{dx}\cos(u(x)) = \frac{d}{du}(\cos u) \cdot \frac{du}{dx} \] \[ = -\sin u \cdot \frac{du}{dx} \]
Chain Rule Formula:
\[ \frac{d}{dx}\cos(u(x)) = -\sin(u(x))\,u'(x) \]
Examples with Solutions
Worked Examples: Differentiating Composite Cosine Functions
Find the derivatives:
- \( f(x) = \cos(2x + 2) \)
- \( g(x) = \cos(\tan x) \)
- \( h(x) = \cos\!\left(\frac{x^2}{x^2+1}\right) \)
Solutions:
-
Let \( u = 2x + 2 \), then \( u' = 2 \).
\[ f'(x) = -2\sin(2x + 2) \] -
Let \( u = \tan x \), then \( u' = \sec^2 x \).
\[ g'(x) = -\sin(\tan x)\sec^2 x \] -
Let \( u = \frac{x^2}{x^2+1} \). Using the quotient rule:
\[ u' = \frac{2x(x^2+1) - x^2(2x)}{(x^2+1)^2} = \frac{2x}{(x^2+1)^2} \] \[ h'(x) = -\frac{2x}{(x^2+1)^2} \sin\!\left(\frac{x^2}{x^2+1}\right) \]