Trig yelp (1 Viewer)

[Run hard@thehsc]

For this question can someone show me the working out as to why the answer is A. Thanks

 

[Flise]

Dunno if there’s an easier way but this is how I did it.

First find the inverse of f(x) (I replaced f(x) with y)
So x = sec(y)

Now differentiate both sides with respect to x
1 = sec(y)tan(y) dy/dx
dy/dx = 1/(sec(y)tan(y))

note: 1 + tan^2(y) = sec^2(y) (make tan(y) the subject)

So tan(y) = square root of (sec^2(y) - 1)

also note we let sec(y) = x

sub into dy/dx

So dy/dx = 1/(x*sqr-root(x^2 - 1))

 

[=)(=]

alternatively differentiate secx get the reciprocal dx/dy which is the derivative of the inverse function

 

[ColdMint123]

Here is an alternative way to do it
Edit: the domain of x should b [0/pi/2]. Also, I have realised that this method is essentially what @Flise has done (their method is more concise as well!)