Question

use implicit differentiation to find \frac{dy}{dx}.
x = \tan y

Explanation:

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

Differentiate \(x\) and \(\tan y\) with respect to \(x\).
$$\frac{d}{dx}(x)=\frac{d}{dx}(\tan y)$$
Since \(\frac{d}{dx}(x) = 1\), and by the chain - rule \(\frac{d}{dx}(\tan y)=\sec^{2}y\frac{dy}{dx}\), we have \(1=\sec^{2}y\frac{dy}{dx}\).

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

Recall the trigonometric identity \(\sec^{2}y = 1+\tan^{2}y\), and since \(x = \tan y\), 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:

\(\frac{dy}{dx}=\frac{1}{1 + x^{2}}\)