Proof of Derivative of sin x
The proof of the derivative of \( \sin (x)\) is presented using the definition of the derivative. The derivative of a sine composite function is also presented including examples with their solutions.
Proof of the Derivative of sin 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 \( f(x) = \sin(x) \) and write the derivative of \( \sin(x) \) as a limit
\( f'(x) = \lim_{h \to 0} \dfrac{\sin(x+h)-\sin(x)}{h} \)
Use the formula \( \sin(x+h) = \sin(x)\cos(h) + \cos(x)\sin(h)\) to rewrite the derivative of \( sin(x) \) as
\( f'(x) = \lim_{h \to 0} \dfrac{\sin(x)\cos(h)+\cos(x)\sin(h)-\sin(x)}{h} \)
Rewrite \( f'(x) \) as follows
\( f'(x) = \lim_{h \to 0} \dfrac{\sin(x) (cos(h) - 1) + \cos(x) \sin(h))}{h} \)
Use the theorem: the limit of the sum of functions is equal to the sum of the limits of these functions to rewrite \( f'(x) \) as follows
\( f'(x) = \lim_{h \to 0} \dfrac{\sin(x) (cos(h) - 1)}{h} + \lim_{h \to 0} \dfrac{\cos(x) \sin(h)}{h} \)
Rewrite the above as
\( f'(x) = \sin(x) \lim_{h \to 0} \dfrac{ (cos(h) - 1)}{h} + \cos(x) \lim_{h \to 0} \dfrac{ \sin(h)}{h} \)
We now use the following results on the limits of trigonometric functions
\( \lim_{h \to 0} \dfrac{\sin(h)}{h} = 1 \) , proved in the use of squeezing theorem to find limits of mathematical functions .
\( \lim_{h \to 0} \dfrac{cos(h) - 1}{h} = 0 \) , proved in calculate limits of trigonometric functions
to simplify \( f'(x) \) to
\( f'(x) = \sin(x) (0) + \cos(x) (1) = cos(x) \)
conclusion
\[ \displaystyle {\dfrac {d}{dx} \sin x = \cos x } \]
Graph of sin x and its Derivative
The graphs of \( \sin(x) \) and its derivative are shown below. Note that at any minimum or maximum of \( \sin(x) \) corresponds a zero of the derivative \( \cos(x) \). Also for any interval over which \( \sin(x) \) is increasing the derivative is positive and for any interval over which \( \sin(x) \) is decreasing, the derivative is negative.
-and-its-derivative.gif)
Derivative of the Composite Function sin(u(x))
Let us consider the composite function sin of another function u(x). Use the chain rule of differentiation to write\( \displaystyle \dfrac{d}{dx} \sin (u(x)) = (\dfrac{d}{du} \sin u) (\dfrac{d}{dx} u ) \)
Simplify
\( = \cos u \dfrac{d}{dx} u \)
Conclusion
\[ \displaystyle \dfrac{d}{dx} \sin (u(x)) = \cos u \dfrac{d}{dx} u \]
Example 1
Find the derivative of the composite sin functions
- \( f(x) = \sin (x^2-x) \)
- \( g(x) = \sin (\sin(x)) \)
- \( h(x) = \sin \left(\dfrac{1-x}{1+x} \right) \)
Solution to Example 1
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Let \( u(x) = x^2-x \) and therefore \( \dfrac{d}{dx} u = \dfrac{d}{dx} (x^2-x) = 2x - 1 \) and apply the rule for the composite sin function given above
\( \displaystyle \dfrac{d}{dx} f(x) = \cos u \dfrac{d}{dx} u = \cos (x^2-x) \times (2x-1) \)
\( = (2x-1) \cos (x^2-x) \)
-
Let \( u(x) = \sin x \) and therefore \( \dfrac{d}{dx} u = \dfrac{d}{dx} \sin x = \cos x \). Apply the above rule of differentiation for the composite sin function
\( \displaystyle \dfrac{d}{dx} g(x) = \cos u \dfrac{d}{dx} u = \cos (\sin x) \times (\cos x) \)
\( = \cos x \cos (\sin x) \)
-
Let \( u(x) = \dfrac{1-x}{1+x} \) and therefore \( \dfrac{d}{dx} u = -\dfrac{2}{(1+x)^2} \) and apply the rule of differentiation for the composite sin function obtained above
\( \displaystyle \dfrac{d}{dx} h(x) = \cos u \dfrac{d}{dx} u = \cos (\dfrac{1-x}{1+x}) \times ( -\dfrac{2}{(1+x)^2}) \)
\( = -\dfrac{2}{(1+x)^2} \cos (\dfrac{1-x}{1+x}) \)
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 .
