How do I differentiate $y= sec^2(x) + tan^2(x)$ ?

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Answer 1 · 2015-03-11T21:53:51

Use the chain rule (generalized power rule) and the derivatives of the trigonometric functions. Beyond that, there are choices you can make

Choice 1:
It may help to re-write it as:
$y=(secx)^2+(tanx)^2$ . So

$(dy)/(dx)=2(secx)*d/(dx)(secx) + 2(tanx)*d/(dx)(tan x)$ Thus<

$(dy)/(dx)=2secx*secxtanx+2tanx*sec^2x$

$(dy)/(dx)=4sec^2xtanx$ .

Which, since $sec^2x=tan^2x+1$ , could also be written

$(dy)/(dx)=4tan^3x+4tanx$

Choice 2:
Use the trigonometric identity to re-write the function using $tanx$ :
$y=sec^2x+tan^2x=(tan^2x+1)+tan^2x=2tan^2x+1$

So,
$(dy)/(dx)=4tanxd/(dx)(tanx)=4tanxsec^2x$

Choice 3:
Use the trigonometric identity just mentioned to re-write the function using $secx$ :

$y=sec^2x+tan^2x=sec^2x+sec^2x-1=2sec^x-1$
So,
$(dy)/(dx)=4secxd/(dx)(secx)=4secxsecxtanx=4sec^2xtanx$

How do I differentiate y= sec^2(x) + tan^2(x)y= sec^2(x) + tan^2(x)?