Proof of Derivative of cos x
The definition of the derivative is used to prove the formula for the derivative of \( \cos (x)\) . The derivative of a cosine composite function is also presented including examples with their solutions.
Proof of the Derivative of cos x Using the Definition
The definition of the derivative \( f' \) of a function \( f \) is given by\[ f'(x) = \lim_{h \to 0} \dfrac{f(x+h)-f(x)}{h} \]
Let \( f(x) = \cos(x) \) and write the derivative of \( \cos(x) \) as a limit
\( f'(x) = \lim_{h \to 0} \dfrac{\cos(x+h)-\cos(x)}{h} \)
Use the formula \( \cos(x+h) = \cos(x)\cos(h) - \sin(x)\sin(h)\) to rewrite the derivative of \( cos(x) \) as
\( f'(x) = \lim_{h \to 0} \dfrac{\cos(x)\cos(h) - \sin(x)\sin(h) - \cos(x)}{h} \)
Rewrite \( f'(x) \) as follows
\( f'(x) = \lim_{h \to 0} \dfrac{\cos(x) (cos(h) - 1) - \sin(x) \sin(h))}{h} \)
Use the theorem on limits that states: the limit of a difference of two functions is equal to the difference of the limits, to rewrite \( f'(x) \) as follows
\( f'(x) = \lim_{h \to 0} \dfrac{\cos(x) (cos(h) - 1)}{h} - \lim_{h \to 0} \dfrac{\sin(x) \sin(h)}{h} \)
Rewrite the above as
\( f'(x) = \cos(x) \lim_{h \to 0} \dfrac{ (cos(h) - 1)}{h} - \sin(x) \lim_{h \to 0} \dfrac{ \sin(h)}{h} \)
We now use the well known results on the limits of trigonometric functions
\( \lim_{h \to 0} \dfrac{\sin(h)}{h} = 1 \) , prooved in the use of squeezing theorem to find limits of mathematical functions .
\( \lim_{h \to 0} \dfrac{cos(h) - 1}{h} = 0 \) , prooved in calculate limits of trigonometric functions
to simplify \( f'(x) \) to
\( f'(x) = \sin(x) (0) - \sin(x) (1) = - sin(x) \)
conclusion
\[ \displaystyle {\dfrac {d}{dx} \cos x = - \sin x } \]
Graph of cos x and its Derivative
The graphs of \( \cos(x) \) and its derivative are shown below. Note that at any maximum or minimum of \( \cos(x) \) corresponds a zero of the derivative. For any interval over which \( \cos(x) \) is increasing the derivative is positive and for any interval over which \( \cos(x) \) is decreasing, the derivative is negative.
-and-its-derivative.gif)
Derivative of the Composite Function cos(u(x))
We now consider the composite function cos of another function u(x). Use the chain rule of differentiation to write\( \displaystyle \dfrac{d}{dx} \cos (u(x)) = (\dfrac{d}{du} \cos u) (\dfrac{d}{dx} u ) \)
Simplify
\( = - \sin u \dfrac{d}{dx} u \)
Conclusion
\[ \displaystyle \dfrac{d}{dx} \cos (u(x)) = - \sin u \dfrac{d}{dx} u \]
Example 1
Find the derivative of the composite cos functions
- \( f(x) = \cos (2x + 2) \)
- \( g(x) = \cos (\tan(x)) \)
- \( h(x) = \cos \left(\dfrac{x^2}{x^2+1} \right) \)
Solution to Example 1
-
Let \( u(x) = 2x+2 \) and therefore \( \dfrac{d}{dx} u = \dfrac{d}{dx} (2x+2) = 2 \) and apply the rule for the composite cos function given above
\( \displaystyle \dfrac{d}{dx} f(x) = \dfrac{d}{dx} \cos(u) = -\sin u \dfrac{d}{dx} u = -\sin (2x+2) \times 2 \)
\( = - 2 \sin (2x+2) \)
-
Let \( u(x) = \tan x \) and therefore \( \dfrac{d}{dx} u = \dfrac{d}{dx} \tan x = \sec^2 x \) and apply the above rule of differentiation for the composite cos function
\( \displaystyle \dfrac{d}{dx} g(x) = \dfrac{d}{dx} \cos(u) = - \sin u \dfrac{d}{dx} u = - \sin (\tan x) \times (\sec^2 x) \)
\( = - \sec^2 x \; \sin (\tan x) \)
-
Let \( u(x) = \left(\dfrac{x^2}{x^2+1} \right) \) and therefore \( \dfrac{d}{dx} u = \frac{2x}{\left(x^2+1\right)^2} \) and apply the rule of differentiation for the composite cos function obtained above
\( \displaystyle \dfrac{d}{dx} h(x) = \dfrac{d}{dx} \cos(u) = - \sin u \dfrac{d}{dx} u = - \sin (\frac{2x}{\left(x^2+1\right)^2}) \times ( \dfrac{2x}{\left(x^2+1\right)^2}) \)
\( = - \dfrac{2x}{\left(x^2+1\right)^2} \cos \left(\dfrac{x^2}{x^2+1} \right) \)
More References and links
derivativedefinition of the derivative
use of squeezing theorem to find limits of mathematical functions .
calculate limits of trigonometric functions
Chain Rule of Differentiation in Calculus .
