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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.

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2 Answers 2

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Hint

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.

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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 $$ {\frac {\mathrm{d}}{\mathrm{d}x}}\left(\int _{a(x)}^{b(x)}f(x,t)\,dt\right) =f\left (x,b(x)\right) \frac {d}{dx}b(x)-f\left(x,a(x)\right){\frac {d}{dx}}a(x)+\int _{a(x)}^{b(x)}{\frac {\partial }{\partial x}}f(x,t)\,dt$$
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