I have the following practice problem that I'd like to know how to solve before taking my test, can someone explain what is necessary?
$D_x \int_{0}^{2x} \left[15 \sqrt{2t^2 + 3t + 4} \right] dt$
I have the following practice problem that I'd like to know how to solve before taking my test, can someone explain what is necessary?
$D_x \int_{0}^{2x} \left[15 \sqrt{2t^2 + 3t + 4} \right] dt$
$\frac{d}{dx}\,\int_{a(x)}^{b(x)}f(t)\,dt = f(b(x))\,b'(x) - f(a(x))\,a'(x)$
In this case, $b(x)=2x$ and $a(x) = 0$.
You know from the fundamental theorem of calculus that $\frac{d}{du}\int_0^uf(t)~dt=f(u)\;.\tag{1}$ If $u=2x$ and $f(t)=15\sqrt{2t^2+3t+4}$, $(1)$ becomes
$\frac{d}{du}\int_0^u15\sqrt{2t^2+3t+4}~dt=15\sqrt{2u^2+3u+4}=15\sqrt{8x^2+6x+4}\;.$
If $F(x)=\int_0^{2x}15\sqrt{2t^2+3t+4}~dt\;,$
you now know $\dfrac{dF}{du}$; how do you get $\dfrac{dF}{dx}$ from this?