Derivative of $x\arctan x$?
I have the following derivative: $$\frac d{dx}x\arctan x$$ I'm unsure of how to solve this derivative; obviously it's a product rule, however is it in a similar vein to finding the derivative of arcsin? Below is my rough idea attempt: $$y=\arctan x\quad\therefore\tan y=x$$ $$\frac d{dx}\tan y=\frac d{dx}x$$ $$\frac{dy}{dx}?=1$$ $$\frac{dy}{dx}=\frac1?$$ Where do I go from here? How do I find $?$ without relying on some "common derivative" knowledge?
Thank you!
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$\begingroup$You don't need to go through the process of "$y = \arctan x$ therefore $\tan y = x$ and..." if you already know the following: $$ \frac d{dx} \arctan x = \frac1{x^2+1}$$
So, you can just use the product rule: $$ \frac d{dx} \left[f(x)g(x)\right] = f'(x) g(x) + f(x)g'(x)$$ with $f(x) = x$ and $g(x) = \arctan x$.
If you don't know that $\dfrac d{dx} \arctan x = \dfrac1{x^2+1}$, then you are on the right track.
$y = \arctan x$, so $x = \tan y$, and then you can differentiate both sides with respect to $x$, making sure to use the chain rule on the right-hand side: \begin{align*} x &= \tan y\\ \frac d{dx} x &= \frac d{dx} \tan y\\ 1 &= (\sec^2 y) \cdot \frac{dy}{dx}\\ \frac1{\sec^2 y} &= \frac{dy}{dx} \end{align*}
Now you just need to express $\sec^2 y$ in terms of $x$. I'll leave that part to you, but here's how to get started: Draw a right triangle and label one of the acute angles and the corresponding sides accordingly so that the picture shows $x = \tan y$. It may be easier to think of it as $\tan y = \dfrac x1 = \dfrac{\text{opposite}}{\text{adjacent}}$.
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