Linda Seales

2021-12-17

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

Raymond Foley

Beginner2021-12-18Added 39 answers

The derivative of $\mathrm{tan}x\text{}\text{is}\text{}{\mathrm{sec}}^{2}x$

To see why, you'll need to know a few results. First, you need to know that the derivative of$\mathrm{sin}x\text{}\text{is}\text{}\mathrm{cos}x$ .

Once all those pieces are in place, the differentiation goes as follows:

$\frac{d}{dx}\mathrm{tan}x$

$=\frac{d}{dx}\frac{\mathrm{sin}x}{\mathrm{cos}x}$

$=\frac{\mathrm{cos}x\cdot \mathrm{cos}x-\mathrm{sin}x\cdot (-\mathrm{sin}x)}{{\mathrm{cos}}^{2}x}\text{}\text{}\text{(using Quotient Rule)}$

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

$=\frac{1}{{\mathrm{cos}}^{2}x}\text{}\text{}\text{(using the Pythagorean Identity)}$

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

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:

Chanell Sanborn

Beginner2021-12-19Added 41 answers

Given, $y=\mathrm{tan}x.$

Let$u=\mathrm{sin}x\text{}\text{and}\text{}v=\mathrm{cos}x$

On applying quotient rule on$y=\frac{\mathrm{sin}x}{\mathrm{cos}x}$ , we get,

$\because \frac{dy}{dx}=\frac{v.d\frac{u}{dx}-u.d\frac{v}{dx}}{{v}^{2}}$

$\Rightarrow \frac{dy}{dx}=\mathrm{cos}x\cdot \mathrm{cos}x-\mathrm{sin}x\frac{-\mathrm{sin}x}{{\mathrm{cos}}^{2}x}$

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

$\Rightarrow \frac{dy}{dx}=\frac{1}{{\mathrm{cos}}^{x}}$

$\Rightarrow \frac{dy}{dx}={\mathrm{sec}}^{2}\left(x\right)$

Thus, the derivative of$y=\mathrm{tan}\left(x\right)\text{}\text{is}\text{}{\mathrm{sec}}^{2}\left(x\right)$

Let

On applying quotient rule on

Thus, the derivative of

RizerMix

Expert2021-12-29Added 600 answers

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