Find dy/dx if y = sec⁻¹(1/2x²-1) differentiation inverse trigonometry ncert problems

`y = sec^(-1)((1)/(2x^(2) -1 )), 0 lt x lt (1)/(sqrt(2))`

Find dy/dx if y = sec^-1(1/(2x²-1)) | Differentiate sec inverse 1 by 2 x square minus 1

y = sec^-1 ( 1 / (2x^2 - 1) ) Find dy/dx | NCERT | CalculusCheck | Class12

Find dy/dx of y=sec^-1(1/2x^2-1), 0 less than x less than 1/√2

Find derivative of y = sec^(-1) (2x +1) with respect to x. Inverse Trig Functions

Q18 | Differentiate sec^(-1)⁡(1/(2x^2-1))

"Find the `(dy)/(dx)`of `y=sec^(-1)(1/(1-2x^2))`"

If `y = sec^(-1) ((1)/(2x^(2) -1)) " then " (dy)/(dx)`= ?

If y= sec inverse[1/2x sq. -1], 0 less than x less than 1/√2, find dy/dx.(+2.M76.HUKAM RAJ BHAGAT.)

Find the Derivative of y = sec(1 + x^2)

Find derivativé of sec⁻¹ (1 / (2x² - 1)) with respect to √(1 - x² ) . shsirclasses

y [cos^-1(2x/1+x^2)],find dy/dx @Mathby1069

Ex 5.3 Q15 Find dy/dx y= sec^-1(1/ (2x^2-1)) || class 12 maths chapter 5 exercise 5.3 question 15

Find dy/dx if y=arcsec(1/(2x^2-1))#shorts

y=sec^-1(1/(2x^2-1)) then dydx || derivative of y = arcsec(1/(2x ^ 2 - 1)) is||

`sec^-1(1/(2x^2-1))`

Derivatives of inverse trigonometric functions sin-1(2x), cos-1 (x^2), tan-1 (x/2) sec-1 (1+x^2)

Derivative of 〖sec〗^(-1) {1/(〖2x〗^2-1)} w.r.t √(1+3x) at x=-1/3 is | DIFFERENTIATION

Find `(dy)/(dx)` in the following: `y=sec^(-1)(1/(2x^2-1))`...