# How do you differentiate  f(x) =1- (sec x/ tan x)^2 ?

May 30, 2018

$\setminus \frac{2 \cos \left(x\right)}{{\sin}^{3} \left(x\right)}$

#### Explanation:

By definition, $\sec \left(x\right) = \setminus \frac{1}{\cos \left(x\right)}$, and $\tan \left(x\right) = \setminus \frac{\sin \left(x\right)}{\cos \left(x\right)}$

We can rewrite the fraction in parenthesis as

$\setminus \frac{\sec \left(x\right)}{\tan \left(x\right)} = \setminus \frac{1}{\cos \left(x\right)} \setminus \cdot \setminus \frac{\cos \left(x\right)}{\sin \left(x\right)} = \setminus \frac{1}{\sin \left(x\right)}$

So, the expression becomes

$1 - \setminus \frac{1}{{\sin}^{2} \left(x\right)}$

To differentiate this expression, remember that

$\frac{d}{\mathrm{dx}} \left(1 - \setminus \frac{1}{{\sin}^{2} \left(x\right)}\right) = \frac{d}{\mathrm{dx}} \left(1\right) - \frac{d}{\mathrm{dx}} \left(\setminus \frac{1}{{\sin}^{2} \left(x\right)}\right) = - \frac{d}{\mathrm{dx}} \setminus \frac{1}{{\sin}^{2} \left(x\right)}$

since the derivative of a number is zero.

Finally, you can write $- \frac{1}{\sin} ^ 2 \left(x\right)$ as $- {\sin}^{- 2} \left(x\right)$, and derive it with chain rule:

$\frac{d}{\mathrm{dx}} - {\sin}^{- 2} \left(x\right) = - \frac{d}{\mathrm{dx}} {\sin}^{- 2} \left(x\right) = - \left(- 2 {\sin}^{- 3} \left(x\right) \cdot \cos \left(x\right)\right) = \setminus \frac{2 \cos \left(x\right)}{{\sin}^{3} \left(x\right)}$

$f ' \left(1 - {\csc}^{2} x\right) = 2 {\csc}^{2} \times x \cot x$

#### Explanation:

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

$\sec x = \frac{1}{\cos} x$
$\tan x = \sin \frac{x}{\cos} x$

$\sec \frac{x}{\tan} x = \frac{\frac{1}{\cos} x}{\sin \frac{x}{\cos} x}$

Multiplying numerator and denominator by cosx

$\sec \frac{x}{\tan} x = \frac{1}{\sin} x$
${\left(\sec \frac{x}{\tan} x\right)}^{2} = {\left(\frac{1}{\sin} x\right)}^{2} = \frac{1}{\sin} ^ 2 x$

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

$f \left(x\right) = 1 - {\csc}^{2} x$

$f ' \left(x\right) = f ' \left(1 - {\csc}^{2} x\right)$

$= f ' \left(1\right) - f ' \left({\csc}^{2} x\right)$

$f ' \left(1\right) = 0$
$f ' \left(\csc 2 x\right) = 2 \csc \times x \left(- \csc \times x \cot x\right)$

$f ' \left({\csc}^{2} x\right) = 2 {\csc}^{2} \times x \cot x$

$f ' \left(1 - {\csc}^{2} x\right) = 0 - \left(- 2 {\csc}^{2} \times x \cot x\right)$

$f ' \left(1 - {\csc}^{2} x\right) = 2 {\csc}^{2} \times x \cot x$