Linda Seales

2021-12-17

What is the derivative of $y=\mathrm{tan}\left(x\right)$?

Raymond Foley

The derivative of
To see why, you'll need to know a few results. First, you need to know that the derivative of .
Once all those pieces are in place, the differentiation goes as follows:
$\frac{d}{dx}\mathrm{tan}x$
$=\frac{d}{dx}\frac{\mathrm{sin}x}{\mathrm{cos}x}$

$=\frac{{\mathrm{cos}}^{2}x+{\mathrm{sin}}^{2}x}{{\mathrm{cos}}^{2}x}$

$={\mathrm{sec}}^{2}x$

Chanell Sanborn

Given, $y=\mathrm{tan}x.$
Let
On applying quotient rule on $y=\frac{\mathrm{sin}x}{\mathrm{cos}x}$, we get,
$\because \frac{dy}{dx}=\frac{v.d\frac{u}{dx}-u.d\frac{v}{dx}}{{v}^{2}}$
$⇒\frac{dy}{dx}=\mathrm{cos}x\cdot \mathrm{cos}x-\mathrm{sin}x\frac{-\mathrm{sin}x}{{\mathrm{cos}}^{2}x}$
$⇒\frac{dy}{dx}=\frac{{\mathrm{cos}}^{2}x+{\mathrm{sin}}^{2}x}{{\mathrm{cos}}^{2}x}$
$⇒\frac{dy}{dx}=\frac{1}{{\mathrm{cos}}^{x}}$
$⇒\frac{dy}{dx}={\mathrm{sec}}^{2}\left(x\right)$
Thus, the derivative of

RizerMix

$y=\mathrm{tan}\left(x\right)$
Differentiate both sides of the equation.
$\frac{d}{dx}\left(y\right)=\frac{d}{dx}\left(\mathrm{tan}\left(x\right)\right)$
The derivative of y with respect to x is y'.
y'
The derivative of $\mathrm{tan}\left(x\right)$ with respect to x is ${\mathrm{sec}}^{2}\left(x\right)$
${\mathrm{sec}}^{2}\left(x\right)$
Reform the equation by setting the left side equal to the right side.
${y}^{\prime }={\mathrm{sec}}^{2}\left(x\right)$
Replace y' with $\frac{dy}{dx}$
$\frac{dy}{dx}={\mathrm{sec}}^{2}\left(x\right)$

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