Question

find the dy/dx if y=sec^-1 (Tanx) ​

Answer 1

Given:

$\star \: y = { \sec}^{ - 1} ( \tan \: x)$

To find :

$\star \: \frac{dy}{dx}$

Solution:

$\implies \: y = { \sec}^{ - 1} ( \tan \: x)$

Differentiate w.rt. x both sides ,

$\implies \: \frac{dy}{dx} = \frac{1}{ | \tan \: x| \sqrt{ { \tan}^{2} x - 1} } \times \frac{d}{dx} \tan \: x \\ \\ \implies \: \: \frac{dy}{dx} = \frac{1}{ | \tan \: x| \sqrt{ { \tan}^{2}x - 1 } } \times { \sec}^{2} x \\ \\ \implies \: \frac{dy}{dx} = \frac{ { \sec}^{2} x}{ | \tan \: x| \sqrt{ { \tan}^{2}x - 1 } }$

Formula:

$\star \bold{ \frac{d}{dx} { \sec}^{ - 1} x = \frac{1}{ |x| \sqrt{ {x}^{2} - 1} }} \\ \\ \star \bold{ \frac{d}{dx} \tan \: x = { \sec}^{2} x}$

Answer 2

Given , the function is

• $\tt y= {sec}^{ - 1} \{tan(x) \}$

Differentiaing y wrt x , we get

$\tt \frac{dy}{dx} = \frac{d}{dx} {sec}^{ - 1} \{tan(x) \}$

$\tt \frac{dy}{dx} = \frac{1}{tan(x) \sqrt{ {tan}^{2} (x) - 1} } \frac{d}{dx} tan(x)$

$\tt \frac{dy}{dx} = \frac{1}{tan(x) \sqrt{ {tan}^{2} (x) - 1} } {sec}^{2} (x)$

$\tt \frac{dy}{dx} = \frac{{sec}^{2} (x)}{tan(x) \sqrt{ {tan}^{2} (x) - 1} }$

Remmember :

$\tt \implies \frac{d}{dx} {sec}^{ - 1} (x) = \frac{1}{x \sqrt{ {(x)}^{2} - 1} }$

$\tt \implies \frac{d}{dx} tan(x) = {sec}^{2}(x)$