Question:
Find the derivative of the function \(f(x) = \sqrt{3x^2 + 2x + 1}\).
Find the derivative of the function \(f(x) = \sqrt{3x^2 + 2x + 1}\).
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For derivatives of square root functions, rewrite as a power and apply the chain rule systematically.
Updated On: May 26, 2025
- $ \frac{6x+2}{2\sqrt{3x^2+2x+1}} $
- $ \frac{3x+1}{\sqrt{3x^2+2x+1}} $
- $ \frac{6x+1}{2\sqrt{3x^2+2x+1}} $
- $ \frac{6x-2}{2\sqrt{3x^2+2x+1}} $
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The Correct Option is A
Solution and Explanation
Rewrite function:
$$
f(x) = (3x^2 + 2x + 1)^{1/2}
$$
Apply chain rule: $$ f'(x) = \frac{1}{2}(3x^2 + 2x + 1)^{-1/2} \cdot (6x + 2) $$
Simplify: $$ f'(x) = \frac{6x + 2}{2\sqrt{3x^2 + 2x + 1}} $$
Apply chain rule: $$ f'(x) = \frac{1}{2}(3x^2 + 2x + 1)^{-1/2} \cdot (6x + 2) $$
Simplify: $$ f'(x) = \frac{6x + 2}{2\sqrt{3x^2 + 2x + 1}} $$
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