How do you differentiate this function?

use this formula :

d(e^u)/du = e^u . du/dx

d(sec u)/dx = secu.tanu. du/dx

so dy/dx = e^sec(2x) . sec2x.tan2x . 2

:p

calr

derivative of exponent*the function

2sec 2x tan 2x*e^sec 2x

y = e^(1 / cos (2x))

y' = (2 *sin (2x) / cos² (2x)) * e^(1 / cos (2x))

<=>

y' = 2 * sin (2x) * sec² (2x) * e^(sec (2x)) =

2 * sec (2x) * tan (2x) * e^(sec (2x))

dy/dx

= e^sec(2x) * d/dx sec(2x)

= e^sec(2x) * sec(2x) tan(2x) * d/dx (2x)

= 2e^sec(2x) * sec(2x) tan(2x).

G(u) = ln(sqrt((5u + 6) / (5u - 6))) G(u) = (a million/2) * ln((5u + 6) / (5u - 6)) G(u) = (a million/2) * (ln(5u + 6) - ln(5u - 6)) G'(u) = (a million/2) * (5 / (5u + 6) - 5 / (5u - 6)) G'(u) = (5/2) * (a million / (5u + 6) - a million / (5u - 6)) G'(u) = (5/2) * ((5u - 6 - 5u - 6)) / (25u^2 - 36) G'(u) = (5/2) * (-12 / (25u^2 - 36)) G'(u) = -30 / (25u^2 - 36)

y = e^sec(2x)

y′ =e^sec(2x) ×d(sec(2x))/dx

=e^sec(2x) ×sec(2x)×tan(2x)×d(2x)/dx

=e^sec(2x) ×sec(2x)×tan(2x)×2

= 2sec(2x)tan(2x)e^sec(2x)