If f(x)=sqrt{x} (x geq 0) and g(x)=1+x^2, the | vedclass

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(D) Given $f(x) = \sqrt{x}$ and $g(x) = 1 + x^2$. The composite function $(f \circ g)(x) = f(g(x)) = \sqrt{1 + x^2}$. To find the derivative $(f \circ

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$\frac{1}{\sqrt{2}}$

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(D) Given $f(x) = \sqrt{x}$ and $g(x) = 1 + x^2$. The composite function $(f \circ g)(x) = f(g(x)) = \sqrt{1 + x^2}$. To find the derivative $(f \circ g)^{\prime}(x)$, we use the chain rule: $(f \circ g)^{\prime}(x) = \frac{d}{dx} (1 + x^2)^{1/2} = \frac{1}{2}(1 + x^2)^{-1/2} \cdot \frac{d}{dx}(1 + x^2)$. $(f \circ g)^{\prime}(x) = \frac{1}{2\sqrt{1 + x^2}} \cdot (2x) = \frac{x}{\sqrt{1 + x^2}}$. Now, substitute $x = 1$ into the derivative: $(f \circ g)^{\prime}(1) = \frac{1}{\sqrt{1 + 1^2}} = \frac{1}{\sqrt{2}}$.

If $f(x)=\sqrt{x}$ $(x \geq 0)$ and $g(x)=1+x^2$, then $(f \circ g)^{\prime}(1)=$

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