expeditiupc

## Answered question

2021-12-31

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

### Answer & Explanation

Mary Nicholson

Beginner2022-01-01Added 38 answers

The derivative of $y=\mathrm{arctan}x$ is ${y}^{\prime }=\frac{1}{1+{x}^{2}}$
We can derive this by using implicit differentiation.
Since inverse tangent is hard to deal with, we rewrite it as $\mathrm{tan}\left(y\right)=x$
By implicitly differentiating with respect to x, ${\mathrm{sec}}^{2}\left(y\right)\cdot {y}^{\prime }=1$
By solbing for y' and using ${\mathrm{sec}}^{2}\left(y\right)=1+{\mathrm{tan}}^{2}\left(y\right)$
${y}^{\prime }=\frac{1}{{\mathrm{sec}}^{2}\left(y\right)}=\frac{1}{{\mathrm{tan}}^{2}\left(y\right)}$
Hence, ${y}^{\prime }=\frac{1}{1+{x}^{2}}$

alkaholikd9

Beginner2022-01-02Added 37 answers

Step 1: Rearrange $y=\mathrm{arctan}\left(x\right)$ as $\mathrm{tan}\left(y\right)=x$.
Step 2: Use implicit differentiation to differentiate this with respect to x, which gives us:
$\left(\frac{dy}{dx}\right)\cdot {\left(\mathrm{sec}\left(y\right)\right)}^{2}=1$
Step 3: Rearrange this equation to give us:
$\frac{dy}{dx}=\frac{1}{{\mathrm{sec}}^{2}\left(y\right)}$

nick1337

Expert2022-01-08Added 777 answers

This helped.

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