Find f' (pi/2) by definition for f(x) = sin x

Find $f'\!\left(\dfrac{\pi}{2}\right)$ by definition for $f(x) = \sin x$.

Step 1: Write the definition of the derivative at $x = \pi/2$

\[ f'\!\left(\frac{\pi}{2}\right) = \lim_{\Delta x \to 0} \frac{f\!\left(\dfrac{\pi}{2} + \Delta x\right) - f\!\left(\dfrac{\pi}{2}\right)}{\Delta x}. \]

Step 2: Compute the increment $\Delta y$

\[ f\!\left(\frac{\pi}{2}+\Delta x\right) = \sin\!\left(\frac{\pi}{2}+\Delta x\right), \qquad f\!\left(\frac{\pi}{2}\right) = \sin\frac{\pi}{2} = 1, \]

so that

\[ \Delta y = \sin\!\left(\frac{\pi}{2}+\Delta x\right) - 1. \]

Step 3: Apply the co-function identity

Since $\sin\!\left(\dfrac{\pi}{2}+\Delta x\right) = \cos(\Delta x)$, the increment simplifies to

\[ \Delta y = \cos(\Delta x) - 1. \]

Step 4: Form the difference quotient and take the limit

\[ \frac{\Delta y}{\Delta x} = \frac{\cos(\Delta x) - 1}{\Delta x}. \]

We use the standard limit

\[ \lim_{u \to 0} \frac{\cos u - 1}{u} = 0, \]

and therefore

\[ f'\!\left(\frac{\pi}{2}\right) = \lim_{\Delta x \to 0} \frac{\cos(\Delta x)-1}{\Delta x} = 0. \]

Step 5: Verification

Using the standard rule $(\sin x)' = \cos x$: \[f'(x) = \cos x \;\Longrightarrow\; f'\!\left(\frac{\pi}{2}\right) = \cos\frac{\pi}{2} = 0 \;\;\checkmark\]

The standard limit itself follows from the half-angle identity $1 - \cos u = 2\sin^{2}\dfrac{u}{2}$: \[\frac{\cos u - 1}{u} = -\frac{2\sin^{2}\dfrac{u}{2}}{u} = -\sin\frac{u}{2}\cdot\frac{\sin\dfrac{u}{2}}{\dfrac{u}{2}} \;\xrightarrow[u \to 0]{}\; -0\cdot 1 = 0.\]