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Find the derivative of f(x) f(x) . f(x)=x+3 f(x) = \sqrt{x+3} \newlinef′(x)= f'(x) = ______

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Q. Find the derivative of f(x) f(x) . f(x)=x+3 f(x) = \sqrt{x+3} \newlinef′(x)= f'(x) = ______
  1. Identify Function: Identify the function to differentiate.\newlinef(x)=x+3f(x) = \sqrt{x+3}\newlineRewrite x+3\sqrt{x+3} as (x+3)12(x+3)^{\frac{1}{2}} for easier differentiation.
  2. Apply Chain Rule: Apply the chain rule for differentiation: ddx[u(v(x))]=u′(v(x))⋅v′(x)\frac{d}{dx}[u(v(x))] = u'(v(x)) \cdot v'(x). Let u(x)=x12u(x) = x^{\frac{1}{2}} and v(x)=x+3v(x) = x + 3. Then, u′(x)=12x−12u'(x) = \frac{1}{2}x^{-\frac{1}{2}} and v′(x)=1v'(x) = 1.
  3. Substitute and Simplify: Substitute back to find f′(x)f'(x).f′(x)=12(x+3)−12×1f'(x) = \frac{1}{2}(x+3)^{-\frac{1}{2}} \times 1Simplify to get f′(x)=12x+3f'(x) = \frac{1}{2\sqrt{x+3}}.

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