Question

Differentiate x. (sec²x)1/x​

Answer 1

Answer:

Thanks for A2A.

We'll go one by one here.

Suppose y=sec²(x)

$Let [math]\displaystyle u=sec(x)$

$\frac{du}{dx}=sec(x)×tan(x)[/math]$

And y becomes

$[math]\displaystyle y=u²$

$\frac{dy}{du} = 2u[/math]$

So from chain rule,

$[math] \displaystyle \frac{dy}{dx}=\frac{dy}{du}×\frac{du}{dx}$

$=2u×sec(x) × tan(x)$

$=2×sec(x)×sec(x)×tan(x)$

$=2sec²(x)tan(x)[/math]$