Derivative of arctan(x/3)

Recognize that $y = \arctan(u)$ where $u = x/3$ is the inner function, so this requires the chain rule. Recall that $\dfrac{d}{dx}[\arctan(u)] = \dfrac{1}{1+u^2}\cdot \dfrac{du}{dx}$. Since $u = x/3$, we have $\dfrac{du}{dx} = \dfrac{1}{3}$.

Substituting $u = x/3$ into the formula gives $\dfrac{dy}{dx} = \dfrac{1}{1+(x/3)^2}\cdot \dfrac{1}{3}$. Simplify $(x/3)^2 = x^2/9$, so the denominator becomes $1 + x^2/9 = \dfrac{9+x^2}{9}$. Combining, $\dfrac{dy}{dx} = \dfrac{1}{3}\cdot \dfrac{9}{9+x^2} = \dfrac{3}{9+x^2}$.

After substituting $u = x/3$ into the arctangent derivative formula, the denominator becomes $\dfrac{9+x^2}{9}$. When we compute $\dfrac{1}{3}\cdot \dfrac{9}{9+x^2}$, we multiply $\dfrac{1}{3}$ by the constant 9 in the numerator, giving the final result $\dfrac{3}{9+x^2}$.