Question

If y = under root( 1- cos2x / 1+ cis 2x), then find its derivative with respect to x, i.e find dy/dx.​

Answer 1

$\huge{ \orange{ \underline{ \overline{ \boxed{ \mathbb{ \blue{ANSWER}}}}}}}$

$\sf{ \green{ \underline{ \underline{GIVEN : }}}}$

$↠ \: \sf \: y \: = \: \sqrt{ \frac{1 \: - \: \cos2x }{1 \: + \: \cos2x} }$

$\sf{ \green{ \underline{ \underline{TO \: FIND : }}}}$

$↠ \sf \: \frac{dy}{dx}$

$\sf{ \green{ \underline{ \underline{SOLUTION : }}}}$

$\implies \: \: \sf \: y \: = \: \sqrt{ \frac{1 \: - \: \cos2x }{1 \: + \: \cos2x} }$

[Using the formula of cos2x and putting the values we get,]

$\implies \: \: \sf \: y \: = \: \sqrt{ \frac{1 \: - \: (1 - 2 {sin}^{2}x) }{1 \: + \: (2 {cos}^{2}x \: - 1) } }$

$\implies \: \: \sf \: y \: = \: \sqrt{ \frac{ \cancel1 \: - \cancel1 \: + \: 2 {sin}^{2}x) }{ \cancel1 \: + \: 2 {cos}^{2}x \: - \: \cancel1 } }$

$\implies \: \: \sf \: y \: = \: \sqrt{ \frac{ 2 {sin}^{2}x}{2 {cos}^{2}x} }$

$\implies \: \: \sf \: y \: = \: \sqrt{ \frac{ \cancel 2 {sin}^{ 2}x}{ \cancel2 {cos}^{2}x} }$

$\implies \: \sf \: y = \sqrt{ { \tan}^{2}x }$

$\boxed{ \therefore \: \: \sf \: y \: = \tan x}$

Now, let us find out derivative of y with respect to x.

$\rightarrow \: \sf \: \frac{d}{dx} (y) \: = \: \frac{d}{dx} (tanx)$

$\boxed{\therefore \: \sf \: \frac{dy}{dx}\: = \: {sec}^{2} x}$

______________________________________

$\sf{ \green{\boxed{FORMULA \: \: USED :}}}$

$↠ \sf{\blue{ \: cos2x \: = \: 1 \: - \: 2 { \sin}^{2}x}}$

$↠ \sf{\blue{ \: cos2x \: = \: 2 { \cos}^{2}x \: - \: 1}}$

$↠ \sf{\blue{ \: \frac{d}{dx}(tanx) = {\sec}^{2} x}}$

#answerwithquality #BAL