# How do you find the derivative of (1-tanx)^2?

Oct 3, 2016

Derivative of ${\left(1 - \tan x\right)}^{2}$ is $- 2 {\sec}^{2} x + 2 \tan x {\sec}^{2} x$

#### Explanation:

We can use Chain rule here. Let $f \left(x\right) = {\left(1 - \tan x\right)}^{2}$. Then we can write it as

$f \left(g \left(x\right)\right) = {\left(g \left(x\right)\right)}^{2}$, where $g \left(x\right) = 1 - \tan x$.

Then $\frac{\mathrm{df}}{\mathrm{dx}} = \frac{\mathrm{df}}{\mathrm{dg}} \times \frac{\mathrm{dg}}{\mathrm{dx}}$

= $2 \times \left(1 - \tan x\right) \times \left(- {\sec}^{2} x\right)$

= $- 2 {\sec}^{2} x + 2 \tan x {\sec}^{2} x$