What is the derivative of arctan(x/3)?

Aug 22, 2017

$\frac{1}{3 + {x}^{2} / 3}$

Explanation:

$\text{differentiate using the "color(blue)"chain rule}$

•color(white)(x)d/dx(arctan(f(x)))=1/(1+(f(x))^2)xxf'(x)

$\Rightarrow \frac{d}{\mathrm{dx}} \left(\arctan \left(\frac{x}{3}\right)\right)$

$= \frac{1}{1 + {x}^{2} / 9} \times \frac{d}{\mathrm{dx}} \left(\frac{1}{3} x\right)$

$= \frac{1}{3 \left(1 + {x}^{2} / 9\right)} = \frac{1}{3 + {x}^{2} / 3}$

Aug 22, 2017

If you haven't memorized the drivative of $\arctan \left(x\right)$. (or you don't trust your memory), see below.

Explanation:

$y = \arctan \left(\frac{x}{3}\right)$

$\tan y = \frac{x}{3}$

Diiferentiate implicitly.

${\sec}^{2} y \frac{\mathrm{dy}}{\mathrm{dx}} = \frac{1}{3}$

$\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{1}{3} {\cos}^{2} \left(y\right)$

Use $\tan y = \frac{x}{3}$ to find $\cos y = \frac{3}{\sqrt{{x}^{2} + 9}}$ (See Note below)

$\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{1}{3} {\left(\frac{3}{\sqrt{{x}^{2} + 9}}\right)}^{2}$

$\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{3}{{x}^{2} + 9}$

Note
There are several possible methods to do this.
I like to draw a right triangle with one angle labeled $y$. The side opposite $y$ is $x$ and the adjacent side is $3$, so ythe hypotenuse is $s q t y \left({x}^{2} + 9\right)$ and the cosine is $\frac{3}{\sqrt{{x}^{2} + 9}}$
Others prefer to use ${\tan}^{2} y + 1 = {\sec}^{2} y$ to find ${\sec}^{2} y$ and the invert.