## Proof

We want to find the derivative of $\mathrm{arctan}(x)$. Let $y=\mathrm{arctan}(x)$. Then, by definition, $x=\mathrm{tan}(y)$.

Taking the derivative of both sides with respect to $x$:

$$\frac{\text{d}}{\text{d}x}(x)=\frac{\text{d}}{\text{d}x}(\mathrm{tan}(y))$$ $$1={\mathrm{sec}}^{2}(y)\xb7\frac{dy}{dx}$$Now, we solve for $\frac{dy}{dx}$:

$$\frac{dy}{dx}=\frac{1}{{\mathrm{sec}}^{2}(y)}$$Using the identity ${\mathrm{sec}}^{2}(y)=1+{\mathrm{tan}}^{2}(y)$ and knowing $\mathrm{tan}(y)=x$:

$$\frac{dy}{dx}=\frac{1}{1+{x}^{2}}$$Thus, the derivative of $\mathrm{arctan}(x)$ is:

$$\frac{\text{d}}{\text{d}x}\mathrm{arctan}(x)=\frac{1}{1+{x}^{2}}$$## Explanation

To find the derivative of $\mathrm{arctan}(x)$, we start by letting $y=\mathrm{arctan}(x)$. This means that $x=\mathrm{tan}(y)$, representing the angle whose tangent is $x$.

Next, we differentiate both sides of the equation $x=\mathrm{tan}(y)$ with respect to $x$. The derivative of $x$ with respect to $x$ is simply $1$.

On the right side, the derivative of $\mathrm{tan}(y)$ with respect to $y$ is ${\mathrm{sec}}^{2}(y)$, and by the chain rule, we multiply by $\frac{dy}{dx}$, which gives us ${\mathrm{sec}}^{2}(y)\xb7\frac{dy}{dx}$.

Setting the derivatives equal, we get:

$$1={\mathrm{sec}}^{2}(y)\xb7\frac{dy}{dx}$$We then solve for $\frac{dy}{dx}$ by dividing both sides by ${\mathrm{sec}}^{2}(y)$:

$$\frac{dy}{dx}=\frac{1}{{\mathrm{sec}}^{2}(y)}$$We use the trigonometric identity ${\mathrm{sec}}^{2}(y)=1+{\mathrm{tan}}^{2}(y)$. Since $\mathrm{tan}(y)=x$ (from our earlier definition), we substitute $x$ for $\mathrm{tan}(y)$:

$${\mathrm{sec}}^{2}(y)=1+{x}^{2}$$Thus, the expression for the derivative simplifies to:

$$\frac{dy}{dx}=\overline{)\frac{1}{1+{x}^{2}}}$$*Q.E.D.*