How do find the derivative of $y = (sec)^2 x + (tan)^2 x$ ?

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Answer 1 · 2015-09-09T09:13:49

Assuming that secx=1/cosx we have that

$y=(1/cosx)^2+(sinx/cosx)^2=>y=(1+(sinx)^2)/(cosx)^2$

Hence the derivative is

$dy/dx=4*tanx*(1/cosx)^2$

How do find the derivative of y = (sec)^2 x + (tan)^2 xy = (sec)^2 x + (tan)^2 x?