How do you find the derivative of ${(1 - tan x)}^{2}$ $(1-tanx)^2$ ?

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Answer 1 · 2016-10-03T09:13:38

Derivative of ${(1 - tan x)}^{2}$ $(1-tanx)^2$ is $- 2 {sec}^{2} x + 2 tan x {sec}^{2} x$ $-2sec^2x+2tanxsec^2x$

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

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

Then $\frac{d f}{d x} = \frac{d f}{d g} \times \frac{d g}{d x}$ $(df)/(dx)=(df)/(dg)xx(dg)/(dx)$

= $2 \times (1 - tan x) \times (- {sec}^{2} x)$ $2xx(1-tanx)xx(-sec^2x)$

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

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