Kathy Williams

2022-01-01

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

Samantha Brown

$\frac{2\mathrm{sin}\left(x\right)\mathrm{cos}\left(x\right)}{{\mathrm{cos}}^{4}\left(x\right)}=2\cdot \frac{\mathrm{sin}\left(x\right)}{\mathrm{cos}\left(x\right)}\cdot \frac{1}{{\mathrm{cos}}^{2}\left(x\right)}=2\mathrm{tan}\left(x\right){\mathrm{sec}}^{2}\left(x\right)$

John Koga

$\frac{2\mathrm{sin}\left(x\right)\mathrm{cos}\left(x\right)}{{\mathrm{cos}}^{2}\left(x\right)\cdot {\mathrm{cos}}^{2}\left(x\right)}=2{\mathrm{sec}}^{2}\left(x\right)\mathrm{tan}\left(x\right)$

Vasquez

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