What is the derivative of $arctan (x - 1)$ $arctan(x-1)$ ?

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Answer 1 · 2016-03-18T19:40:31

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

We must first know the derivative of $arctan (x)$ $arctan(x)$ , which is:

$\frac{d}{d x} arctan (x) = \frac{1}{1 + x^{2}}$ $d/dxarctan(x)=1/(1+x^2)$

According to the chain rule, we see that

$d/dxarctan(f(x))=1/(1+(f(x))^2)*f'(x)$

So, for $arctan(x-1)$ , we see that $f(x)=x-1$ and $f'(x)=1$ .

$d/dxarctan(x-1)=1/(1+(x-1)^2)*1$

$=1/(1+(x^2-2x+1)$

$=1/(x^2-2x+2)$

What is the derivative of arctan(x-1)arctan(x−1)arctan(x-1)?