Find dydx if y-sec-1-12x2-1- 0 1-2

y= ^ 1 ( 1 2 x ^ 2 1 #160;) #160; x=cosθ #160; 2 x ^ 2 1=2cos ^ 2 θ #160; 1=cos 2θ #160; y= ^ 1 ( 1 cos 2θ #160; #16 ...

You visited us 3 times! Enjoying our articles? Unlock Full Access!

Principal Solution of Trigonometric Equation

Find dydx i...

Question

Find $\frac{d y}{d x}$ if $y = s e c^{- 1} [\frac{1}{2 x^{2} - 1}]$ , $0 < x > \frac{1}{\sqrt{2}}$

Open in App

Solution

Suggest Corrections

Similar questions

y = (s e c^{- 1} x)^{2}, x > 0

\frac{d y}{d x}

\frac{d y}{d x} = \frac{1 - cos x}{1 + cos x}

\frac{d y}{d x} = \sqrt{4 - y^{2}} (- 2 < y < 2)

\frac{d y}{d x} = 1 (y \neq 1)

{sec}^{2} x tan y d x + {sec}^{2} y tan x d y = 0

(e^{x} + e^{- x}) d y - (e^{x} - e^{- x}) d x = 0

\frac{d y}{d x} = (1 + x^{2}) (1 + y^{2})

y log y d x - x d y = 0

x^{4} \frac{d y}{d x} = - y^{5}

\frac{d y}{d x} = {sin}^{- 1} x

e^{x} tan y d x + (1 - e^{x}) {sec}^{2} y d y = 0

f^{'} (x) = \sqrt{2 x^{2} - 1}

y = f (x^{2})

\frac{d y}{d x}

x = 1

\frac{d y}{d x}

y = {sec}^{- 1} (\frac{1}{2 x^{2} - 1}), 0 < x < \frac{1}{\sqrt{2}}

e^{x} tan y d x + (1 - e^{x}) {sec}^{2} y d y = 0

Join BYJU'S Learning Program