Derivative of Integral -

It's true that if $F(x)=\int_a^x f(t)\,dt$, then $F'(x)=f(x)$. But in your case you have $$ h(x)=F(\tfrac{10}x), $$ where $F(x)=\int_2^x\arctan(t)\,dt$. So to find $h'$ you need to apply the Chain Rule and the Fundamental Theorem of Calculus.

if $F(x) - F(b) = \int_b^{x} f(t) \ dt$

$F'(x) = f(x)$

This is the fundamental theorem of calculus.

We have

$F(\frac {10}{x}) - F(b) = \int_b^{\frac{10}{x}} f(t) \ dt$

$\frac {d}{dx}\left(F(\frac {10}{x}) - F(b)\right) = \frac {d}{dx}\int_b^{\frac{10}{x}} f(t) \ dt$

Remember, b is a constant, so $\frac {d}{dx} F(b) = 0$

For the other term we are going to use the chain rule.

$F'(\frac {10}{x}) = f(\frac {10}{x})(\frac {d}{dx} \frac {10}{x}) = f(\frac{10}{x})(\frac {-10}{x^2})$

$-\frac {10\arctan \frac {10}{x}}{x^2}$