what is the derivative of \sec^2 x I'm using the q

what is the derivative of $$\displaystyle{{\sec}^{{2}}{x}}$$
I'm using the quotient rule and so I rewrite $$\displaystyle{{\sec}^{{2}}{x}}\ \text{ as }\ {\frac{{{1}}}{{{{\cos}^{{2}}{x}}}}}$$
Then set $$\displaystyle{u}={1},{u}′={0}\ \text{ then }\ {v}={{\cos}^{{2}}{x}}$$ and $$\displaystyle{v}′=-{2}{\sin{{x}}}{\cos{{x}}}$$
I then get $$\displaystyle{\frac{{{2}{\sin{{x}}}{\cos{{x}}}}}{{{{\cos}^{{4}}{x}}}}}$$
but when I simplify that I don't get $$\displaystyle{2}{{\sec}^{{2}}{x}}{\tan{{x}}}$$ which is the correct answer?

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Samantha Brown
$$\displaystyle{\frac{{{2}{\sin{{\left({x}\right)}}}{\cos{{\left({x}\right)}}}}}{{{{\cos}^{{4}}{\left({x}\right)}}}}}={2}\cdot{\frac{{{\sin{{\left({x}\right)}}}}}{{{\cos{{\left({x}\right)}}}}}}\cdot{\frac{{{1}}}{{{{\cos}^{{2}}{\left({x}\right)}}}}}={2}{\tan{{\left({x}\right)}}}{{\sec}^{{2}}{\left({x}\right)}}$$
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John Koga
$$\displaystyle{\frac{{{2}{\sin{{\left({x}\right)}}}{\cos{{\left({x}\right)}}}}}{{{{\cos}^{{2}}{\left({x}\right)}}\cdot{{\cos}^{{2}}{\left({x}\right)}}}}}={2}{{\sec}^{{2}}{\left({x}\right)}}{\tan{{\left({x}\right)}}}$$
Vasquez

$$\begin{array}{}\frac{2 \sin x \cos x}{\cos^4 x}= \\=\frac{2(\sin x)}{(\cos x)[\cos^2 x]}= \\=2\tan x \sec^2 x \end{array}$$