▶️ Answer/Explanation
Solution
Correct Answer: A
Step 1: Recall the derivative of arctan(u):
\[ \frac{d}{dx} \arctan(u) = \frac{u’}{1 + u^2} \]
Step 2: Apply to our function \( y = \arctan(4x) \):
\[ \frac{dy}{dx} = \frac{4}{1 + (4x)^2} \]
Step 3: Evaluate at \( x = \frac{1}{4} \):
\[ \frac{dy}{dx}\bigg|_{x=\frac{1}{4}} = \frac{4}{1 + (4 \times \frac{1}{4})^2} = \frac{4}{1 + 1} = \frac{4}{2} = 2 \]
Explanation:
- Use the chain rule for the derivative of inverse tangent.
- Substitute \( u = 4x \) and compute the derivative.
- Plug in the given x-value and simplify to find the slope.
▶️ Answer/Explanation
Solution
Correct Answer: D
Step 1: Recall the derivative formula for inverse tangent:
\[ \frac{d}{dx} \tan^{-1}(u) = \frac{u’}{1 + u^2} \]
Step 2: Apply to our function with \( u = 3x \):
\[ \frac{d}{dx} \tan^{-1}(3x) = \frac{3}{1 + (3x)^2} \]
Explanation:
- The derivative of \(\tan^{-1}(u)\) is \(\frac{u’}{1+u^2}\).
- Here, \(u = 3x\), so \(u’ = 3\).
- Substitute into the formula to get the correct answer.
Why others are incorrect:
- A) \(\sec^2\) is the derivative of \(\tan\), not \(\tan^{-1}\)
- B) \(\csc^2\) is unrelated to inverse tangent
- C) This would be correct for \(\sin^{-1}\), not \(\tan^{-1}\)
▶️ Answer/Explanation
Solution
Correct Answer: C
Step 1: Recognize the limit as the definition of the derivative:
\[ \lim_{x\rightarrow \frac{1}{2}} \frac{\arcsin x – \arcsin(\frac{1}{2})}{x-\frac{1}{2}} = f’\left(\frac{1}{2}\right) \]
Step 2: Find the derivative of \( \arcsin x \):
\[ f'(x) = \frac{1}{\sqrt{1-x^2}} \]
Step 3: Evaluate at \( x = \frac{1}{2} \):
\[ f’\left(\frac{1}{2}\right) = \frac{1}{\sqrt{1-\left(\frac{1}{2}\right)^2}} = \frac{1}{\sqrt{\frac{3}{4}}} = \frac{2}{\sqrt{3}} = \frac{2\sqrt{3}}{3} \]
Note: The answer simplifies to \( \frac{2}{\sqrt{3}} \), which is equivalent to \( \frac{2\sqrt{3}}{3} \). Explanation:
- The given limit represents the derivative of \( \arcsin x \) at \( x = \frac{1}{2} \).
- The derivative of \( \arcsin x \) is a standard result.
- Substitution and simplification leads to the final answer.
Why others are incorrect:
- A) The derivative is not zero at this point
- B) This is the value of \( \arcsin(\frac{1}{2}) \), not its derivative
- D) The derivative exists at \( x = \frac{1}{2} \)
▶️ Answer/Explanation
Solution
Correct Answer: C
Step 1: Recognize the limit represents the derivative of arcsin at point a:
\[ f'(a) = \frac{1}{\sqrt{1-a^2}} \] Step 2: The derivative exists only when:
\[ 1 – a^2 > 0 \implies -1 < a < 1 \] Step 3: Evaluate each option:
- A) \(\frac{\sqrt{2}}{2} \approx 0.707\) → valid
- B) \(\frac{\sqrt{3}}{2} \approx 0.866\) → valid
- C) \(\sqrt{3} \approx 1.732\) → invalid (outside domain)
- D) \(\frac{1}{2} = 0.5\) → valid
Conclusion: Option C (\(\sqrt{3}\)) is the value that could not be a, as it lies outside the domain where the derivative exists.