Let f(x)=∫xx21+t4 dtf(x) = \int_{x}^{x^2} \sqrt{1+t^4} \, dtf(x)=∫xx21+t4dt. Find the value of f′(1)f'(1)f′(1).
f′(1)=2f'(1) = \sqrt{2}f′(1)=2
f′(1)=0f'(1) = 0f′(1)=0
f′(1)=22−2=2f'(1) = 2\sqrt{2} - \sqrt{2} = \sqrt{2}f′(1)=22−2=2
f′(1)=32f'(1) = 3\sqrt{2}f′(1)=32