# How do you find the derivative of tan(x − y) = x?

Jul 15, 2015

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

#### Explanation:

I'm assuming you want to find $\frac{\mathrm{dy}}{\mathrm{dx}}$. For this we first need an expression for $y$ in terms of $x$. We note that this problem has various solutions, since $\tan \left(x\right)$ is a periodic functions, $\tan \left(x - y\right) = x$ will have multiple solutions. However, since we know the period of the tangent function ($\pi$), we can do the following: $x - y = {\tan}^{- 1} x + n \pi$, where ${\tan}^{- 1}$ is the inverse function of the tangent giving values between $- \frac{\pi}{2}$ and $\frac{\pi}{2}$ and the factor $n \pi$ has been added to account for the periodicity of the tangent.

This gives us $y = x - {\tan}^{- 1} x - n \pi$, therefore $\frac{\mathrm{dy}}{\mathrm{dx}} = 1 - \frac{d}{\mathrm{dx}} {\tan}^{- 1} x$, note that the factor $n \pi$ has disappeared. Now we need to find $\frac{d}{\mathrm{dx}} {\tan}^{- 1} x$. This is quite tricky, but doable using the reverse function theorem.

Setting $u = {\tan}^{- 1} x$, we have $x = \tan u = \sin \frac{u}{\cos} u$, so $\frac{\mathrm{dx}}{\mathrm{du}} = \frac{{\cos}^{2} u + {\sin}^{2} u}{\cos} ^ 2 u = \frac{1}{\cos} ^ 2 u$, using the quotient rule and some trigonometric identities. Using the inverse function theorem (which states that if $\frac{\mathrm{dx}}{\mathrm{du}}$ is continuous and non-zero, we have $\frac{\mathrm{du}}{\mathrm{dx}} = \frac{1}{\frac{\mathrm{dx}}{\mathrm{du}}}$), we have $\frac{\mathrm{du}}{\mathrm{dx}} = {\cos}^{2} u$. Now we need to express ${\cos}^{2} u$ in terms of x.

To do this, we use some trigonometry. Given a right triangle with sides $a , b , c$ where $c$ is the hypotenuse and $a , b$ connected to the right angle. If $u$ is the angle where side $c$ intersects side $a$, we have $x = \tan u = \frac{b}{a}$. With the symbols $a , b , c$ in the equations we denote de length of these edges. $\cos u = \frac{a}{c}$ and using Pythagoras theorem, we find $c = \sqrt{{a}^{2} + {b}^{2}} = a \sqrt{1 + {\left(\frac{b}{a}\right)}^{2}} = a \sqrt{1 + {x}^{2}}$. This gives $\cos u = \frac{1}{\sqrt{1 + {x}^{2}}}$, so $\frac{\mathrm{du}}{\mathrm{dx}} = \frac{1}{1 + {x}^{2}}$.

Since $u = {\tan}^{- 1} x$, we can substitute this into our equation for $\frac{\mathrm{dy}}{\mathrm{dx}}$ and find $\frac{\mathrm{dy}}{\mathrm{dx}} = 1 - \frac{1}{1 + {x}^{2}} = {x}^{2} / \left(1 + {x}^{2}\right)$.