Find (phi'(1))/(f'(1)) given f(x) = 1-x and phi(x) = 1
Find $\dfrac{\varphi'(1)}{f'(1)}$ given $f(x) = 1-x$ and $\varphi(x) = 1-\sin\dfrac{\pi x}{2}$.
Explanation of Method
A ratio of slopes at a common zero.
Step 1: Differentiate $f$
\[ f(x) = 1-x \;\Longrightarrow\; f'(x) = -1, \qquad f'(1) = -1. \]
Step 2: Differentiate $\varphi$
Chain rule with inner function $\dfrac{\pi x}{2}$, whose derivative is $\dfrac{\pi}{2}$: $\varphi'(x) = -\cos\frac{\pi x}{2}\cdot\frac{\pi}{2} = -\frac{\pi}{2}\cos\frac{\pi x}{2}$.
At $x = 1$ the cosine vanishes: \[\varphi'(1) = -\frac{\pi}{2}\cos\frac{\pi}{2} = -\frac{\pi}{2}\cdot 0 = 0.\]
Step 3: Form the ratio
\[ \frac{\varphi'(1)}{f'(1)} = \frac{0}{-1} = 0. \]
Step 4: Verification
\[f(1) = 0 \quad\text{and}\quad \varphi(1) = 1-\sin\frac{\pi}{2} = 1-1 = 0\]
\[\frac{\varphi(1+10^{-4})-\varphi(1-10^{-4})}{2\times 10^{-4}} \approx 0 \;\;\checkmark\]
Both functions vanish at $x = 1$: they meet the $x$-axis at the same point. There $f$ crosses with slope $-1$, while $\varphi$ is tangent to the axis.
Result
Final answer:
\[\boxed{\,\dfrac{\varphi'(1)}{f'(1)} = 0\,}\]
Remark
Remark. The ratio of derivatives at a common zero is exactly what L'Hôpital's rule computes: since $f(1) = \varphi(1) = 0$, it follows that $\displaystyle\lim_{x\to 1}\frac{\varphi(x)}{f(x)} = \frac{\varphi'(1)}{f'(1)} = 0$. So $\varphi$ vanishes at $x = 1$ to higher order than $f$ — second order, in fact, since $\varphi''(1) = \tfrac{\pi^{2}}{4} \neq 0$.