▶️ Answer/Explanation
Solution
For \( g(x) = 3\sin x + 2\cos x + 5 \), the derivative is \( g'(x) = 3\cos x – 2\sin x \).
Evaluate at \( x = \frac{\pi}{3} \): \( g’\left(\frac{\pi}{3}\right) = 3\cos\left(\frac{\pi}{3}\right) – 2\sin\left(\frac{\pi}{3}\right) = 3 \cdot \frac{1}{2} – 2 \cdot \frac{\sqrt{3}}{2} = \frac{3}{2} – \sqrt{3} \).
✅ Answer: A)
▶️ Answer/Explanation
Solution
The expression \( \lim_{h \to 0} \frac{\cos(x + h) – \cos x}{h} \) is the derivative of \( g(x) = \cos x \), so \( g'(x) = -\sin x \).
Evaluate at \( x = \frac{\pi}{6} \): \( g’\left(\frac{\pi}{6}\right) = -\sin\left(\frac{\pi}{6}\right) = -\frac{1}{2} \).
✅ Answer: C)
▶️ Answer/Explanation
Solution
The limit \( \lim_{h \to 0} \frac{7e^x – 7e^{x+h}}{4h} \) simplifies to \( \frac{7e^x – 7e^x e^h}{4h} = \frac{7e^x (1 – e^h)}{4h} \).
As \( h \to 0 \), this is the derivative of \( f(x) = -7e^x \) divided by 4: \( f'(x) = -7e^x \), so \( \frac{f'(x)}{4} = \frac{-7e^x}{4} \).
✅ Answer: C)
▶️ Answer/Explanation
Solution
For \( f(x) = e^{1/x} \), use the chain rule. Let \( u = \frac{1}{x} \), so \( f(x) = e^u \), and \( \frac{du}{dx} = -\frac{1}{x^2} \).
Then, \( f'(x) = e^u \cdot \frac{du}{dx} = e^{1/x} \cdot \left(-\frac{1}{x^2}\right) = -\frac{e^{1/x}}{x^2} \).
✅ Answer: A)