# Finding the derivative of integral function of the form $\int_{x^2}^{x^4}(\int_{t^2}^{t^4}g(y)dy)dt$

I need to find the derivative of the following function on [1,$$\infty$$): $$f(x)=\int_{x^2}^{x^4}\left(\int_{t^2}^{t^4}\frac{1}{1+\sqrt{y}}dy\right)dt$$ My intuition is that $$h(y)=\dfrac{1}{1+\sqrt{y}}$$ is continuous on every interval $$[a,b]$$ so $$G(t)=\displaystyle \int_{t^2}^{t^4}\frac{1}{1+\sqrt{y}}dy$$ is the antiderivative and $$G'(t)=h(t)$$, but then I have a problem with the next step of finding the connection with $$f'(x)$$. also, I'm not sure that what I just wrote is correct.

Any help will be appreciated.

Set $$F(x)=\int_0^x G(t)\,\mathrm d t.$$ Then $$f(x)=F(x^4)-F(x^2).$$ Using the chain rule knowing that $$F'(x)=G(x)$$ allows you to conclude.
The main idea is this: $$f(x)=\int_{x^2}^{x^4}\int_{t^2}^{t^4}\dfrac{1}{1+\sqrt{y}}\mathrm{d}y\mathrm{d}t\\ \sqrt{y}=q, \quad y=q^2,\quad \mathrm{d}y=2q\mathrm{d}q\\ \int_{x^2}^{x^4}\int_{t}^{t^2}\dfrac{2q}{1+q}\mathrm{d}q\mathrm{d}t\\ \int_{x^2}^{x^4}\left.2\left(q-\ln(q+1)\right)\right|_{q=t}^{q=t^2}\mathrm{dt}\\ \int_{x^2}^{x^4}2\left(t^2-\ln(t^2+1)-(t-\ln(t+1))\right)\mathrm{dt}\\ \int_{x^2}^{x^4}2\left(t^2-t-\ln(t^2+1)+\ln(t+1))\right)\mathrm{dt}\\ \int_{x^2}^{x^4}2\left(t^2-t-\ln(t^2+1)+\ln\left(t+1\right))\right)\mathrm{dt}\\$$ At this point you can do 2 things:
1. Evaluate the 4 integrals in $$t=x^4$$ and $$t=x^2$$ and then you calculate the derivative
2. You can use Leibniz integral rule $${\frac {\mathrm{d}}{\mathrm{d}x}}\left(\int _{a(x)}^{b(x)}f(x,t)\,dt\right) =f\left (x,b(x)\right) \frac {d}{dx}b(x)-f\left(x,a(x)\right){\frac {d}{dx}}a(x)+\int _{a(x)}^{b(x)}{\frac {\partial }{\partial x}}f(x,t)\,dt$$