Answer :
By the fundamental theorem of calculus,
[tex]\displaystyle\frac{\mathrm d}{\mathrm dx}\int_0^x\cos^{-1}t\,\mathrm dt=\cos^{-1}x[/tex]
so that
[tex]f'(0.3)=\cos^{-1}0.3\approx1.266[/tex]
By the fundamental theorem of calculus,
[tex]\displaystyle\frac{\mathrm d}{\mathrm dx}\int_0^x\cos^{-1}t\,\mathrm dt=\cos^{-1}x[/tex]
so that
[tex]f'(0.3)=\cos^{-1}0.3\approx1.266[/tex]