Given: $$F(x) = \int_{2-x^{0.5}}^{x^4} \frac{\arctan(y/x)}{y}dy$$ Calculate $F'(4)$ or ($lim x-> 4$ if $F'(4)$ not defined)
I was given that if: $G(x)=\int_{u(x)}^{v(x)} g(x,y)dy$ then $G(x_0)=\int_{u(x_0)}^{v(x_0)} g(x_0,y)dy$
In my case I get: $F(4) = \int_{0}^{4^4} \frac{\arctan(y/4)}{y}dy$ So, $F'(4) = \left(\int_{0}^{4^4} \frac{\arctan(y/4)}{y}dy\right)'$
I tried using Leibniz's rule but I get that the derivative is always o,
which is wrong. What's my mistake and May someone correct me?
$G(x)=\int_u^vg(x,y)dy=h(x,v)-h(x,u)$, so $G'(x)=\frac{\partial h(x,v)}{\partial x}+\frac{\partial h(x,v)}{\partial v}\frac{dv}{dx}-\frac{\partial h(x,u)}{\partial x}-\frac{\partial h(x,u)}{\partial u}\frac{du}{dx}=\int_u^v\frac{\partial g(x,y)}{\partial x}dy+g(v,x)\frac{dv}{dx}-g(u,x)\frac{du}{dx}$
I'll let you finish.