Find the derivative of f given by f(x) = tan^ | vedclass

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(A) Let $y = 	an^{-1} x$. Then,$x = 	an y$. Differentiating both sides with respect to $x$,we get: $1 = \sec^2 y \cdot \frac{dy}{dx}$ This implies t

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$\frac{1}{1+x^2}$

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(A) Let $y = 	an^{-1} x$. Then,$x = 	an y$. Differentiating both sides with respect to $x$,we get: $1 = \sec^2 y \cdot \frac{dy}{dx}$ This implies that: $\frac{dy}{dx} = \frac{1}{\sec^2 y}$ Using the trigonometric identity $\sec^2 y = 1 + 	an^2 y$,we have: $\frac{dy}{dx} = \frac{1}{1 + 	an^2 y}$ Since $x = 	an y$,we substitute $x$ back into the equation: $\frac{dy}{dx} = \frac{1}{1 + x^2}$

Find the derivative of $f$ given by $f(x) = \tan^{-1} x$,assuming it exists.

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