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

Jun 6, 2015

Final Answer: $\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{- {x}^{2}}{1 + {x}^{2}}$

Let $z = \left(x + y\right)$

$\tan \left(x + y\right) = x \setminus \quad \setminus \rightarrow \setminus \quad \tan \left(z\right) = x$

Take the derivative of each side with respect to x

$\frac{d}{\mathrm{dx}} \left[\tan \left(z\right)\right] = \frac{d}{\mathrm{dx}} \left[x\right]$

${\sec}^{2} \left(z\right) \frac{\mathrm{dz}}{\mathrm{dx}} = 1 \setminus \quad \setminus \leftarrow \setminus \quad \setminus \textrm{\left(C h a \in R \underline{e}\right)}$

Now find what $\frac{\mathrm{dz}}{\mathrm{dx}}$ is

$\frac{\mathrm{dz}}{\mathrm{dx}} = \frac{d}{\mathrm{dx}} \left[x + y\right] = \frac{d}{\mathrm{dx}} \left[x\right] + \frac{d}{\mathrm{dx}} \left[y\right] = 1 + \frac{\mathrm{dy}}{\mathrm{dx}}$

substitute $z$ and $\frac{\mathrm{dz}}{\mathrm{dx}}$ into: $\setminus \quad {\sec}^{2} \left(z\right) \frac{\mathrm{dz}}{\mathrm{dx}} = 1$ and get

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

Solve for $\frac{\mathrm{dy}}{\mathrm{dx}}$

$\frac{\mathrm{dy}}{\mathrm{dx}} = {\cos}^{2} \left(x + y\right) - 1 = - {\sin}^{2} \left(x + y\right)$

I used the trig identity: ${\cos}^{2} \left(z\right) - 1 = - {\sin}^{2} \left(z\right)$ in the last step.

Substitute $y$ to get the answer purely in terms of $x$

$\tan \left(x + y\right) = x \setminus \quad \setminus \rightarrow \setminus \quad y = \arctan \left(x\right) - x$

$\frac{\mathrm{dy}}{\mathrm{dx}} = - {\sin}^{2} \left(\arctan \left(x\right)\right)$

Answer can be simplified with some work
Let $\theta = \arctan \left(x\right) \setminus \implies \tan \left(\theta\right) = \frac{x}{1} = \setminus \frac{\textrm{o p p o s i t e}}{\setminus} \textrm{a \mathrm{dj} a c e n t}$

The above expression implies the triangle below

Sub $\arctan \left(x\right) = \theta$ into the $\frac{\mathrm{dy}}{\mathrm{dx}}$ expression

$\frac{\mathrm{dy}}{\mathrm{dx}} = - {\sin}^{2} \left(\theta\right) = - {\left(\setminus \frac{\textrm{o p p o s i t e}}{\setminus} \textrm{h y p o t e \nu s e}\right)}^{2}$

From the diagram we know $\sin \left(\theta\right) = \frac{x}{\setminus} \sqrt{1 + {x}^{2}}$. So...

$\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{- {x}^{2}}{1 + {x}^{2}}$