Question

Let \( f(1) = 3 \), \( f'(1) = -\frac{1}{3} \), \( g(1) = -4 \), and \( g'(1) = -\frac{8}{3} \). The derivative of \( \sqrt{[f(x)]^2 + [g(x)]^2} \) w.r.t. \( x \) at \( x = 1 \) is ........

• A. \( \frac{-29}{25} \)
• B. \( \frac{7}{3} \)
• C. \( \frac{31}{15} \)
• D. \( \frac{29}{15} \)

Hint:

For derivatives of composite functions, use the chain rule and simplify step by step.

Answer 1

The Correct Option is A

Step 1: Apply the chain rule.
We are tasked with differentiating \( \sqrt{[f(x)]^2 + [g(x)]^2} \). Using the chain rule: \[ \frac{d}{dx}\left( \sqrt{[f(x)]^2 + [g(x)]^2} \right) = \frac{1}{2\sqrt{[f(x)]^2 + [g(x)]^2}} \cdot \left( 2f(x)f'(x) + 2g(x)g'(x) \right) \] Simplifying: \[ \frac{d}{dx} = \frac{f(x)f'(x) + g(x)g'(x)}{\sqrt{[f(x)]^2 + [g(x)]^2}} \]

Step 2: Substitute the given values.
At \( x = 1 \), we have \( f(1) = 3 \), \( f'(1) = -\frac{1}{3} \), \( g(1) = -4 \), and \( g'(1) = -\frac{8}{3} \). Substituting these values into the formula: \[ \frac{3 \times \left(-\frac{1}{3}\right) + (-4) \times \left(-\frac{8}{3}\right)}{\sqrt{3^2 + (-4)^2}} = \frac{-1 + \frac{32}{3}}{\sqrt{9 + 16}} = \frac{-1 + \frac{32}{3}}{5} = \frac{-\frac{3}{3} + \frac{32}{3}}{5} = \frac{29}{15} \cdot \frac{1}{5} = \frac{-29}{25} \]

Step 3: Conclude.
Thus, the correct answer is \( \frac{-29}{25} \).

Final Answer: \[ \boxed{\frac{-29}{25}} \]