## Calculus (3rd Edition)

(b) $\alpha'(x)= A(x)\alpha(x).$
Since $\alpha(x)=e^{\int P(x)dx}$, then $$\alpha'(x)=P(x)e^{\int P(x)dx}=P(x)\alpha(x).$$ Then the right answer is: (b) $\alpha'(x)= A(x)\alpha(x).$