## Calculus: Early Transcendentals (2nd Edition)

$$- \left( {{{\cos }^4}x + 6} \right)\sin x$$
\eqalign{ & \frac{d}{{dx}}\int_0^{\cos x} {\left( {{t^4} + 6} \right)dt} \cr & {\text{using The Fundamental Theorem }}\left( {{\text{part 1}}} \right) \cr & A\left( x \right) = \int_a^x {f\left( t \right)dt{\text{ }} \to {\text{ }}} A'\left( x \right) = \frac{d}{{dx}}\int_a^x {f\left( t \right)dt} = f\left( x \right) \cr & {\text{the upper limit is }}{x^2},{\text{ so is needed to apply the chain rule}} \cr & \frac{d}{{dx}}\int_0^{\cos x} {\left( {{t^4} + 6} \right)dt} = \frac{{dy}}{{dx}} = \frac{{dy}}{{du}}\frac{{du}}{{dx}} \cr & {\text{with }}u = \cos x \cr & = \left( {\frac{d}{{du}}\int_0^u {\left( {{t^4} + 6} \right)dt} } \right)\frac{d}{{dx}}\left( {\cos x} \right) \cr & = \left( {\frac{d}{{du}}\int_0^u {\left( {{t^4} + 6} \right)dt} } \right)\left( { - \sin x} \right) \cr & {\text{by the fundamental theorem }}\left( {{\text{part 1}}} \right) \cr & = \left( {{u^4} + 6} \right)\left( { - \sin x} \right) \cr & = \left( {{{\left( {\cos x} \right)}^4} + 6} \right)\left( { - \sin x} \right) \cr & = - \left( {{{\cos }^4}x + 6} \right)\sin x \cr}