Question

(sec^-1)(1 + (tan^2x))/(1 - (tan^2x))
differentiate w.r.t. x

Answer 1

For a while, let \(u=1+\tan^2x\).
\[y=\frac{\sec^{-1}u}{2-u}\\
y'=\frac{(2-u)\frac{1}{u\sqrt{u^2-1}}+\sec^{-1}u}{(2-u)^2}\cdot \frac{du}{dx}\]

Answer 2

lol yea i got it i changed\[\sec ^{-1}\] to \[\cos ^{-1}\] and then used the trigonometric formula for cos2x