How derivative of (1/cot(x)) = sec^2(x) ?

May 10, 2018

$\frac{d}{\mathrm{dx}} \frac{1}{\cot} x = \frac{d}{\mathrm{dx}} t g x = {\sec}^{2} x$ because, $t g x = \sin \frac{x}{\cos} x$, like this you have to derive using the rule of the quotient. See bellow:

Explanation:

$\frac{d}{\mathrm{dx}} t g x = \frac{d}{\mathrm{dx}} \sin \frac{x}{\cos} x = \frac{\sin x ' \cos x - \sin x \cos x '}{\cos x} ^ 2 = \frac{\cos x \cos x - \sin x \left(- \sin x\right)}{\cos x} ^ 2 = \frac{{\cos}^{2} x + {\sin}^{2} x}{\cos x} ^ 2 = \frac{1}{\cos} ^ 2 x = {\sec}^{2} x$.

I hope to helped you!

May 10, 2018

${\tan}^{2} x + 1 = 2 {\sec}^{2} x \cdot \tan x$

Explanation:

show below

$\frac{d}{\mathrm{dx}} \left[\frac{1}{\cot} x\right] = \frac{d}{\mathrm{dx}} \left[{\sec}^{2} x\right]$

$\cot x = \cos \frac{x}{\sin} x$

$\frac{d}{\mathrm{dx}} \left[\sin \frac{x}{\cos} x\right] = \frac{d}{\mathrm{dx}} \left[{\sec}^{2} x\right]$

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

$\frac{{\cos}^{2} x + {\sin}^{2} x}{\cos} ^ 2 x = 2 {\sec}^{2} x \cdot \tan x$

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

${\tan}^{2} x + 1 = 2 {\sec}^{2} x \cdot \tan x$