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:
\[ f'(x) = \lim_{h \to 0} \dfrac{\sin(x)\cos(h)+\cos(x)\sin(h)-\sin(x)}{h} \]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:
\[ f'(x) = \lim_{h \to 0} \dfrac{\sin(x) (\cos(h) - 1)}{h} + \lim_{h \to 0} \dfrac{\cos(x) \sin(h)}{h} \]Factor out terms independent of \( h \):
\[ f'(x) = \sin(x) \lim_{h \to 0} \dfrac{\cos(h) - 1}{h} + \cos(x) \lim_{h \to 0} \dfrac{\sin(h)}{h} \]We now use the following standard trigonometric limits:
\[ \lim_{h \to 0} \dfrac{\sin(h)}{h} = 1 \](proved via the Squeezing Theorem)
\[ \lim_{h \to 0} \dfrac{\cos(h) - 1}{h} = 0 \](proved in calculating limits of trigonometric functions)
Substitute these limits into the expression for \( f'(x) \):
\[ f'(x) = \sin(x)(0) + \cos(x)(1) = \cos(x) \]Graph of sin(x) and Its Derivative
The graphs of \( \sin(x) \) and its derivative are shown below. Note that any minimum or maximum of \( \sin(x) \) corresponds to a zero of the derivative \( \cos(x) \). Additionally, over any interval where \( \sin(x) \) is increasing, the derivative is positive, and where \( \sin(x) \) is decreasing, the derivative is negative.
-and-its-derivative.gif)
Derivative of the Composite Function sin(u(x))
Let us consider the composite function sine of another function \( u(x) \). Use the chain rule of differentiation to write:
\[ \dfrac{d}{dx} \sin(u(x)) = \left(\dfrac{d}{du} \sin u\right) \left(\dfrac{d}{dx} u\right) \] \[ = \cos u \dfrac{du}{dx} \]Examples with Solutions
Worked Examples: Differentiating Composite Sine Functions
Find the derivative of the composite sine functions:
- \( f(x) = \sin(x^2 - x) \)
- \( g(x) = \sin(\sin x) \)
- \( h(x) = \sin \left(\dfrac{1-x}{1+x}\right) \)
Solutions:
-
Let \( u(x) = x^2 - x \), then \( \dfrac{du}{dx} = 2x - 1 \).
Applying the rule for the composite sine function: \[ f'(x) = \cos(u) \dfrac{du}{dx} = \cos(x^2 - x) \cdot (2x - 1) = (2x - 1) \cos(x^2 - x) \] -
Let \( u(x) = \sin x \), then \( \dfrac{du}{dx} = \cos x \).
Applying the rule: \[ g'(x) = \cos(u) \dfrac{du}{dx} = \cos(\sin x) \cdot \cos x = \cos x \cos(\sin x) \] - Let \( u(x) = \dfrac{1-x}{1+x} \). Using the quotient rule: \[ \dfrac{du}{dx} = \dfrac{(-1)(1+x) - (1-x)(1)}{(1+x)^2} = \dfrac{-1-x-1+x}{(1+x)^2} = \dfrac{-2}{(1+x)^2} \] Applying the rule: \[ h'(x) = \cos(u) \dfrac{du}{dx} = \cos \left(\dfrac{1-x}{1+x}\right) \cdot \left(-\dfrac{2}{(1+x)^2}\right) = -\dfrac{2}{(1+x)^2} \cos \left(\dfrac{1-x}{1+x}\right) \]