# How do you implicitly differentiate tan(x+y) = x?

Dec 28, 2015

$\frac{d \left(y\right)}{\mathrm{dx}} = \left(1 - {\sec}^{2} \left(x + y\right)\right) \left({\cos}^{2} \left(x + y\right)\right)$$\ldots \ldots \ldots . . \left(i\right)$
$\frac{d \left(x\right)}{\mathrm{dy}} = \frac{1}{\left(1 - {\sec}^{2} \left(x + y\right)\right) \left({\cos}^{2} \left(x + y\right)\right)}$$\ldots \ldots \ldots \ldots \ldots \left(i i\right)$

#### Explanation:

$\tan \left(x + y\right) = x$
Regard $y$ as a function of x and differentiate with respect to $x$.
$\implies \frac{d \left(\tan \left(x + y\right)\right)}{\mathrm{dx}} = \frac{d \left(x\right)}{\mathrm{dx}}$
using chain rule
$\implies {\sec}^{2} \left(x + y\right) \left(\frac{d \left(x\right)}{\mathrm{dx}} + \frac{d \left(y\right)}{\mathrm{dx}}\right) = 1$
$\implies {\sec}^{2} \left(x + y\right) \left(1 + \frac{d \left(y\right)}{\mathrm{dx}}\right) = 1$
$\implies \frac{d \left(y\right)}{\mathrm{dx}} = \frac{1}{\sec} ^ 2 \left(x + y\right)$
$\implies \frac{d \left(y\right)}{\mathrm{dx}} = \frac{1}{\sec} ^ 2 \left(x + y\right) - 1$
$\implies \frac{d \left(y\right)}{\mathrm{dx}} = \frac{1 - {\sec}^{2} \left(x + y\right)}{\sec} ^ 2 \left(x + y\right)$
$\implies \frac{d \left(y\right)}{\mathrm{dx}} = \left(1 - {\sec}^{2} \left(x + y\right)\right) \left({\cos}^{2} \left(x + y\right)\right)$$\ldots \ldots \ldots . . \left(i\right)$
$\implies \frac{d \left(x\right)}{\mathrm{dy}} = \frac{1}{\left(1 - {\sec}^{2} \left(x + y\right)\right) \left({\cos}^{2} \left(x + y\right)\right)}$$\ldots \ldots \ldots \ldots \ldots \left(i i\right)$

$\left(i\right)$ represents the derivative of $y$ with respect to $x$ and $\left(i i\right)$ represents the derivative of $x$ with respect to $y$.