5.3 The Fundamental Theorem of Calculus/13

$h(x)=\int _{2}^{1/x}\arctan(t)dt$

Failed to parse (syntax error): {\displaystyle \frac{d}{dx}\left[h(x)\right] =\frac{d}{dx}\left[\int_{2}^{1/x}\arctan(t)dt\right] =\frac{-1}{x^2}\cdot(\arctan\left(\frac{1}{x}\right))-0\cdot(\arctan\left(2)\right) =\frac{-\arctan\frac{1}{x}}

${\text{Therefore, }}g'(x)=$