Answer:
[tex]4\sqrt{16x^2+1}[/tex]
Step-by-step explanation:
Given [tex]\frac{d}{dx} \int\limits^{4x}_ {1} \, \sqrt{t^2+1}\ dt[/tex]
Using Fundamental Theorem of Calculus
[tex]\frac{d}{dx} \int\limits^{x}_ {a} \, f(t)dt =f(x)\ \ \ for\ any\ constant\ a[/tex]
[tex]\frac{d}{dx} \int\limits^{4x}_ {1} \, \sqrt{t^2+1}\ dt=\frac{d(4x)}{dx}\frac{d}{dx}\int\limits^{4x}}_{1}\, \sqrt{t^2+1}dt[/tex]
Now, [tex]\frac{d(4x)}{dx}=4\ and\ \frac{d}{d(4x)} \int\limits^{4x}_ {1} \, \sqrt{t^2+1}\ dt=\sqrt{(4x^2)+1}[/tex]
Hence plugging these results we get:
[tex]\frac{d}{dx} \int\limits^{4x}_ {1} \, \sqrt{t^2+1}\ dt=4\sqrt{(4x)^2+1}\\\\=4\sqrt{16x^2+1}[/tex]
