Find the derivative of f(x)=∫x3x2t2+1dtf(x) = \int_{x^3}^{x^2} \sqrt{t^2+1} dtf(x)=∫x3x2t2+1dt.
2xx4+1−3x2x6+12x\sqrt{x^4+1} - 3x^2\sqrt{x^6+1}2xx4+1−3x2x6+1
x4+1−x6+1\sqrt{x^4+1} - \sqrt{x^6+1}x4+1−x6+1
x2x4+1−x3x6+1x^2\sqrt{x^4+1} - x^3\sqrt{x^6+1}x2x4+1−x3x6+1
3x2x4+1−2xx6+13x^2\sqrt{x^4+1} - 2x\sqrt{x^6+1}3x2x4+1−2xx6+1