Question

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

Explanation:

Step1: Differentiate both sides

Differentiate $x = \tan y$ with respect to $x$. The derivative of $x$ with respect to $x$ is $1$, and for $\tan y$ with respect to $x$ we use the chain - rule. Let $u = y$, so $\frac{d}{dx}(\tan y)=\sec^{2}y\frac{dy}{dx}$. We get $1=\sec^{2}y\frac{dy}{dx}$.

Step2: Solve for $\frac{dy}{dx}$

Isolate $\frac{dy}{dx}$ by dividing both sides of the equation $1=\sec^{2}y\frac{dy}{dx}$ by $\sec^{2}y$. So $\frac{dy}{dx}=\frac{1}{\sec^{2}y}$. Since $\cos^{2}y=\frac{1}{\sec^{2}y}$, then $\frac{dy}{dx}=\cos^{2}y$.

Answer:

$\cos^{2}y$