Find $dy/dx$ for $y =\operatorname{arcsec}(\tan x)$

Question

If $$y =\operatorname{arcsec}(\tan x),$$ then find $\frac{{dy}}{{dx}}$.

Note: I was practising on finding derivatives for various types of questions and I stumbled upon the question mentioned above. I solved it and my answer came out to be : $$\frac{{(\sec x)^2}}{{\tan x \cdot \tan(\operatorname{arcsec}(\tan x))}}$$

I couldn't move further after this expression.

However, the options provided for this question were :

a) $$\frac{x}{\sqrt{1+x^2}}$$

b) $$-\frac{x}{\sqrt{1+x^2}}$$

c) $$\frac{x}{\sqrt{1-x^2}}$$

d) none

And the correct option mentioned in the book is option (a). I don't seem to understand how option (a) is given as the answer to this question.

Please help me in understanding how option (a) will be the solution. Is the answer different to option (a) ?

Thank you.

Answer 1

I think in order the get correct option $$y=sec(\arctan(x))$$ $$tan(z)=x$$ Using pythagorean theorem $$sec(z)=\sqrt{x^2+1}$$ $$z=arcsec(\sqrt{x^2+1})$$ $$z=arctan{x}$$ So $$y=sec(z)$$ $$y=sec(arcsec(\sqrt{x^2+1})$$ $$y=\sqrt{x^2+1}$$ $$y'=\frac{x}{\sqrt{x^2+1}}$$