Let f(x) = tan^{-1} x. Then f'(x) + f''(x) = | vedclass

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(B) Given $f(x) = 	an^{-1} x$. First derivative: $f'(x) = \frac{1}{1+x^2}$. Second derivative: $f''(x) = \frac{d}{dx} (1+x^2)^{-1} = -1(1+x^2)^{-2} \

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$1$

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(B) Given $f(x) = 	an^{-1} x$. First derivative: $f'(x) = \frac{1}{1+x^2}$. Second derivative: $f''(x) = \frac{d}{dx} (1+x^2)^{-1} = -1(1+x^2)^{-2} \cdot (2x) = \frac{-2x}{(1+x^2)^2}$. We are given $f'(x) + f''(x) = 0$. Substituting the derivatives: $\frac{1}{1+x^2} - \frac{2x}{(1+x^2)^2} = 0$. Multiply by $(1+x^2)^2$: $(1+x^2) - 2x = 0$. This simplifies to $x^2 - 2x + 1 = 0$, which is $(x-1)^2 = 0$. Therefore, $x = 1$.

Let $f(x) = \tan^{-1} x$. Then $f'(x) + f''(x) = 0$ when $x$ is equal to:

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