Are you referring to $\int_{0}^{x}f(t)w(t)dt$ and $\int f(x)w(x)dx$? Well, if we differentiate the first and second, we arrive at $f(x)w(x)$, but for conceptually different reasons.
We know that $\frac{d}{dx}\int_{0}^{x}f(t)w(t)dt=f(x)w(x)$
due to the following proof (part of the Fundamental Theorem of Calculus):
Let $u(t)=f(t)w(t)$. Define $U(t)=\int u(t)dt$. From this, we have $ \begin{align} \frac{d}{dx}\int_0^{x}f(t)w(t)dt&=\frac{d}{dx}\int_0^{x}u(t)dt\\ &=\frac{d}{dx}\left[ \int u(t)dt\right]^{x}_{0}\\ &=\frac{d}{dx}\left[U(x)-U(0)\right]\\ &=\frac{d}{dx}U(x)+\frac{d}{dx}U(0)\\ &=\frac{d}{dx}U(x)+0\\ &=u(x)\\ &=f(x)w(x). \end{align} $
We know that $\frac{d}{dx}\int f(x)w(x)dx=f(x)w(x)$ on the basis of the fact that differentiation and integration are inverse operations just like addition and subtraction. They cancel one another, in other words.
Attempt To Address Your Concerns More Aptly
Let's assume the author means $y=\exp\left(\int_0^{x}f(t)w(t)dt\right).$
Make the $u$ sub as above and say more aptly $y=\exp\left(\int_0^{x}u(t)dt\right).$
When we differentiate this, we must apply the chain rule: $\frac{dy}{dx}=\frac{dy}{dv}\frac{dv}{dx}.$
Let $v=\int_0^{x}u(t)dt$. We then have:
$ \begin{align} \frac{dy}{dx}&=\exp(v)\frac{d}{dx}v\\ &=\exp\left({\int_0^x u(t)dt}\right)\frac{d}{dx}\int_0^{x}u(t)dt\\ &=\exp\left({\int_0^x u(t)dt}\right)u(x)\\ &=\exp\left(\int_0^{x}f(t)w(t)dt\right)f(x)w(x). \end{align}$
Second Derivative
$ \begin{align} y''&=\left[\exp\left({\int_0^x u(t)dt}\right)u(x) \right]'\\ &=\left(\exp\left({\int_0^x u(t)dt}\right)\right)'u(x)+\exp\left({\int_0^x u(t)dt}\right)u'(x) \quad (\text{product rule: } (fg)'=f'g+fg'.)\\ &=\exp\left(\int_0^{x}u(t)dt\right)u(x)u(x)+\exp\left({\int_0^x u(t)dt}\right)u'(x)\\ &=\exp\left(\int_0^{x}u(t)dt\right)u^2(x)+\exp\left({\int_0^x u(t)dt}\right)u'(x)\\ &=\exp\left(\int_0^{x}u(t)dt\right)u^2(x)+\exp\left({\int_0^x u(t)dt}\right)u'(x)\\ &=\exp\left(\int_0^{x}f(t)w(t)dt\right)f^2(x)w^2(t)+\exp\left({\int_0^x f(t)w(t)dt}\right)(f(t)w(t))'\\ &=\exp\left(\int_0^{x}f(t)w(t)dt\right)f^2(x)w^2(t)+\exp\left({\int_0^x f(t)w(t)dt}\right)(f(t)'w(t)+f(t)w'(t)). \end{align} $