▶️ Answer/Explanation
Solution
Given \( 3x – \tan y = 4 \), differentiate implicitly with respect to \( x \): \( \frac{d}{dx}(3x) – \frac{d}{dx}(\tan y) = \frac{d}{dx}(4) \).
This gives \( 3 – \sec^2 y \cdot \frac{dy}{dx} = 0 \).
Solve for \( \frac{dy}{dx} \): \( \sec^2 y \cdot \frac{dy}{dx} = 3 \), so \( \frac{dy}{dx} = \frac{3}{\sec^2 y} \).
Since \( \sec^2 y = \frac{1}{\cos^2 y} \), then \( \frac{dy}{dx} = 3 \cos^2 y \).
✅ Answer: B)
▶️ Answer/Explanation
Solution
Rewrite \( \cot x = \frac{\cos x}{\sin x} \). Use the quotient rule: let \( u = \cos x \), \( v = \sin x \), so \( u’ = -\sin x \), \( v’ = \cos x \).
The quotient rule gives: \( \frac{d}{dx}(\cot x) = \frac{u’v – uv’}{v^2} = \frac{(-\sin x)(\sin x) – (\cos x)(\cos x)}{(\sin x)^2} = \frac{-\sin^2 x – \cos^2 x}{\sin^2 x} = \frac{-(\sin^2 x + \cos^2 x)}{\sin^2 x} = \frac{-1}{\sin^2 x} = -\csc^2 x \).
✅ Answer: D)
▶️ Answer/Explanation
Solution
Note that \( \csc x = \frac{1}{\sin x} \), so \( \sin x \csc x = \sin x \cdot \frac{1}{\sin x} = 1 \) for \( \sin x \neq 0 \).
The derivative of a constant is 0. Thus, \( \frac{d}{dx}(\sin x \csc x) = 0 \).
✅ Answer: A)
▶️ Answer/Explanation
Solution
Step 1: \( \csc x \cdot \sin x = 1 \) — Correct, since \( \csc x = \frac{1}{\sin x} \), so \( \csc x \cdot \sin x = 1 \).
Step 2: \( \frac{d}{dx}(\csc x \cdot \sin x) = 0 \) — Correct, as the derivative of a constant (1) is 0.
Step 3: \( \frac{d}{dx}(\csc x) \cdot \sin x + \csc x \cdot \cos x = 0 \) — Correct, using the product rule on \( \csc x \cdot \sin x \), then setting equal to 0.
Step 4: \( \frac{d}{dx}(\csc x) = \frac{-\csc x \cdot \cos x}{\sin x} = -\csc x \cdot \cot x \) — Correct, solving for the derivative and simplifying using \( \cot x = \frac{\cos x}{\sin x} \).
Since all steps are correct, there is no error.
✅ Answer: D)