Question

use implicit differentiation to find \\( \frac { d y } { d x } \\).
\\( x = \tan y \\)
\\( \frac { d y } { d x } = \square \\)

Explanation:

Step1: Differentiate both sides with respect to \(x\)

Differentiate \(x\) with respect to \(x\) gives \(1\). For \(\tan y\), using the chain - rule \((\tan u)^\prime=\sec^{2}u\cdot u^\prime\) (where \(u = y\) and \(u^\prime=\frac{dy}{dx}\)), we have \(\frac{d}{dx}(\tan y)=\sec^{2}y\frac{dy}{dx}\). So the equation becomes \(1=\sec^{2}y\frac{dy}{dx}\).

Step2: Solve for \(\frac{dy}{dx}\)

Since \(\sec^{2}y = 1+\tan^{2}y\) and \(x = \tan y\), then \(\sec^{2}y=1 + x^{2}\). From \(1=\sec^{2}y\frac{dy}{dx}\), we can solve for \(\frac{dy}{dx}\) by dividing both sides by \(\sec^{2}y\). So \(\frac{dy}{dx}=\frac{1}{\sec^{2}y}\). Substituting \(\sec^{2}y = 1 + x^{2}\), we get \(\frac{dy}{dx}=\frac{1}{1 + x^{2}}\).

Answer:

\(\frac{1}{1 + x^{2}}\)