What's the derivative of $f (x) = arctan [\frac{x}{\sqrt{5 - x^{2}}}]$ $f(x) = arctan[x/(sqrt(5-x^2))]$ ?

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Answer 1 · 2017-02-18T09:17:32

$\frac{1}{\sqrt{5 - x^{2}}}$ $1/sqrt(5-x^2)$

Let $x = \sqrt{5} sin t$ $x = sqrt5sint$ . Then,

$f = arctan (tan t) = t = arcsin (\frac{1}{\sqrt{5}} x)$ $f =arctan(tant)=t=arcsin(1/sqrt5x)$ . So,

$f'=1/sqrt(1-x^2/5)(x/sqrt5)'$

$=sqrt5/(sqrt(5-x^2))(1/sqrt5)=1/sqrt(5-x^2)$

What's the derivative of f(x) = arctan[x/(sqrt(5-x^2))]f(x)=arctan[x√5−x2] f(x) = arctan[x/(sqrt(5-x^2))]?