What is the derivative of $y = cot (arctan x)$ $y=cot (arctan x)$ ?

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Answer 1 · 2017-09-13T10:15:31

$\frac{d}{d x} cot (arctan x) = - \frac{1}{x^{2}}$ $d/dx cot (arctan x) = -1/x^2$

We have:

$y = cot (arctan x)$ $y=cot (arctan x)$

Suppose we let:

$tan u = x \Leftrightarrow u = arctan x$ $tanu = x <=> u = arctan x$

Then:

$cot u = \frac{1}{tan u} = \frac{1}{x}$ $cot u = 1/(tan u) = 1/x$

But, we established that $u = arctan x$ $u = arctan x$ , therefore:

$cot (arctan x) = \frac{1}{x}$ $cot (arctan x) = 1/x$

Hence, we have:

$y = \frac{1}{x} \Rightarrow \frac{d y}{d x} = - \frac{1}{x^{2}}$ $y = 1/x => dy/dx = -1/x^2$

What is the derivative of y=cot (arctan x) y=cot(arctanx) y=cot (arctan x) ?