`y = tanh^-1(sin(2x))` Find the derivative of the function

This is a composite function, and to differentiate it we need the chain rule,

`(f(g(x)))' = f'(g(x))*g'(x).`

Here  `g(x) = sin(2x)`  and ` f(z) = tanh^-1(z),` so we need their derivatives also. They are known, `(sin(2x))' = 2cos(2x)`  (we use the chain rule for `2x` here),  `(tanh^-1(z))' = 1/(1 - z^2).`

This way the result is  `y'(x) = 1/(1 - sin^2(2x))*2cos(2x),`

which is equal to `(2cos(2x))/(cos^2(2x)) = 2/(cos(2x)) = 2sec(2x).`

This is a composite function, and to differentiate it we need the chain rule,

`(f(g(x)))' = f'(g(x))*g'(x).`

Here  `g(x) = sin(2x)`  and ` f(z) = tanh^-1(z),` so we need their derivatives also. They are known, `(sin(2x))' = 2cos(2x)`  (we use the chain rule for `2x` here),  `(tanh^-1(z))' = 1/(1 - z^2).`

This way the result is  `y'(x) = 1/(1 - sin^2(2x))*2cos(2x),`

which is equal to `(2cos(2x))/(cos^2(2x)) = 2/(cos(2x)) = 2sec(2x).`