Jamir Summers

2022-11-26

Find the derivative of $y={\mathrm{tan}}^{2}\left(x\right)$?

Melanie Wong

Expert

$\frac{dy}{dx}=\frac{d}{dx}\left({\mathrm{tan}}^{2}\left(x\right)\right)$
$y=f\left(g\left(x\right)\right)$
$\frac{df\left(g\left(x\right)\right)}{dx}=\frac{d}{d\left(g\left(x\right)\right)}f\left(g\left(x\right)\right)×\frac{d}{dx}g\left(x\right)$
Here $f\left(g\left(x\right)\right)={\left[\mathrm{tan}\left(x\right)\right]}^{2}$ and $g\left(x\right)=\mathrm{tan}\left(x\right)$
$\frac{d}{dx}\left(\mathrm{tan}{\left(x\right)\right)}^{2}=\frac{d}{d\left(\mathrm{tan}\left(x\right)\right)}\left(\mathrm{tan}{\left(x\right)\right)}^{2}×\frac{d}{dx}\left(\mathrm{tan}\left(x\right)\right)$
$\frac{d}{dx}\left({\mathrm{tan}}^{2}\left(x\right)\right)=2\mathrm{tan}\left(x\right)×\frac{d}{dx}\left(\mathrm{tan}x\right)\phantom{\rule{0ex}{0ex}}\frac{dy}{dx}=2\mathrm{tan}\left(x\right)×{\mathrm{sec}}^{2}\left(x\right)\left[\frac{d}{dx}\mathrm{tan}\left(x\right)={\mathrm{sec}}^{2}\left(x\right)\right]\phantom{\rule{0ex}{0ex}}\frac{dy}{dx}=2{\mathrm{sec}}^{2}\left(x\right)\mathrm{tan}\left(x\right)$
Hence, the derivative of $y={\mathrm{tan}}^{2}\left(x\right)$ is $\frac{dy}{dx}=2{\mathrm{sec}}^{2}\left(x\right)\mathrm{tan}\left(x\right)$

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