23. tan(x + y) = x (0, 0)

Explanation:

Step1: Differentiate both sides

Differentiate $\tan(x + y)$ with respect to $x$ using the chain - rule. The derivative of $\tan(u)$ with respect to $x$ is $\sec^{2}(u)\cdot\frac{du}{dx}$, where $u=x + y$. So the derivative of $\tan(x + y)$ with respect to $x$ is $\sec^{2}(x + y)\left(1+\frac{dy}{dx}\right)$, and the derivative of $x$ with respect to $x$ is $1$. So we have $\sec^{2}(x + y)\left(1+\frac{dy}{dx}\right)=1$.

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

First, expand the left - hand side: $\sec^{2}(x + y)+\sec^{2}(x + y)\frac{dy}{dx}=1$. Then, isolate $\frac{dy}{dx}$: $\sec^{2}(x + y)\frac{dy}{dx}=1 - \sec^{2}(x + y)$. Since $1+\tan^{2}\theta=\sec^{2}\theta$, we know that $1 - \sec^{2}(x + y)=-\tan^{2}(x + y)$. So $\frac{dy}{dx}=\frac{1 - \sec^{2}(x + y)}{\sec^{2}(x + y)}=\frac{-\tan^{2}(x + y)}{\sec^{2}(x + y)}$.

Step3: Evaluate at the point $(0,0)$

Substitute $x = 0$ and $y = 0$ into the derivative. Since $\tan(0 + 0)=0$ and $\sec(0 + 0)=1$, we have $\frac{dy}{dx}\big|_{(0,0)}=\frac{-\tan^{2}(0)}{\sec^{2}(0)}=0$.

Answer:

$0$