Find $H'(2)$ given that $H(x) = \int^{x^3-4}_{2x} \frac{x}{1+\sqrt{t}}dt$

Question

Find $H'(2)$ given that $$H(x) = \int^{x^3-4}_{2x} \frac{x}{1+\sqrt{t}}dt$$

I can break up the following function using the union integral property. Let $c \in \mathbb{R}$.

$$H(x) = \int^{c}_{2x} \frac{x}{1+\sqrt{t}}dt + \int^{x^3-4}_{c} \frac{x}{1+\sqrt{t}}dt$$

Since I need a form, $H(x) = \int^{x}_{a} h(t) dt$, I can rewrite the first integral

$$H(x) = -\int^{2x}_{c} \frac{x}{1+\sqrt{t}}dt + \int^{x^3-4}_{c} \frac{x}{1+\sqrt{t}}dt$$

This is a combination of functions. Thus,

$$H(x) = -H(2x) + H(x^3-4)$$ $$H'(x) = -H'(2x) + H'(x^3-4)$$

I'm not sure about the last two steps, are they correct? After I find $H'(x)$ I can find $H'(2)$

Answer 1

differentiating after the Leibniz rule we get $$\int_{2\,x}^{{x}^{3}-4}\! \left( 1+\sqrt {t} \right) ^{-1}\,{\rm d}t+3 \,{\frac {{x}^{3}}{1+\sqrt {{x}^{3}-4}}}-2\,{\frac {x}{1+\sqrt {2} \sqrt {x}}} $$ plugging $$x=2$$ in the formel above we obtain $$\frac{20}{3}$$

Answer 2

A problem like this is best solved with Leibniz integration rule for differentiating under an integral. It states that

$${\frac {\mathrm {d} }{\mathrm {d} x}}\left(\int _{a(x)}^{b(x)}f(t,x)\,\mathrm {d} t\right)=\int _{a(x)}^{b(x)}{\frac {\partial f}{\partial x}}\,\mathrm {d} t\,+\,f{\big (}b(x),x{\big )}\cdot b'(x)\,-\,f{\big (}a(x),x{\big )}\cdot a'(x)$$

which is easily done with $f(x,t)=\frac x{1+\sqrt t}$, $a(x)=2x$, and $b(x)=x^3-4$.

As to your last steps, you forgot to apply chain rule.