# Thread: Trigonometry Derivative ('Show That' Question)

1. ## Trigonometry Derivative ('Show That' Question)

Q: Given that $\displaystyle y= \mathrm{arctan} \frac{x}{a}$, where $\displaystyle a$ is a constant, show that $\displaystyle x \frac{\mathrm{d}y}{\mathrm{d}x} = \frac{\tan y}{1 + \tan ^2 y}$.

So... I have done (and reached a dead end):

$\displaystyle y= \mathrm{arctan} \frac{x}{a}$
$\displaystyle \therefore \tan y = \frac {x}{a}$
$\displaystyle \frac{\tan y}{1 + \tan ^2 y}$
$\displaystyle = \frac{(\frac{x}{a})}{1 + (\frac{x}{a})^2}$
$\displaystyle = \frac{(\frac{x}{a})}{\frac{a^2 + x^2}{a^2}}$
$\displaystyle =\frac{ax}{a^2 + x^2}$

2. $\displaystyle y = arctan(\frac{x}{a})$

From the table of standard integrals, we know that

$\displaystyle \int \frac{1}{a^2+b^2}\,dx = \frac{1}{a}arctan(\frac{x}{a})$
But $\displaystyle \frac{1}{a}= 1$ in our equation. Therefore a = 1 (not too sure about this step. I think if you went forward and worked with x/a it will cancel out anyway)

$\displaystyle y = arctan(\frac{x}{a})$
$\displaystyle \frac{dy}{dx} = \frac{1}{1^2+ x^2}$
Multiplying both sides by x
$\displaystyle x\frac{dy}{dx} = \frac{x}{1+x^2}$
But $\displaystyle x = tan (y)$
$\displaystyle x\frac{dy}{dx} = \frac{tan (y)}{1+ tan^2(y)}$