\begin{array}{rcl}
\displaystyle\frac{\psi(x+h)-\psi(x)}{h}&=&\displaystyle\frac 1h\left(\int_{x_0}^{x+h}K(x+h,\alpha)f(\alpha)\,d\alpha-\int_{x_0}^{x}K(x,\alpha)f(\alpha)\,d\alpha\right)
\vspace{0.5em}\\&=&\displaystyle
\frac 1h\left(\int_{x_0}^{x+h}K(x+h,\alpha)f(\alpha)\,d\alpha-\int_{x_0}^{x+h}K(x,\alpha)f(\alpha)\,d\alpha\right)+\frac 1h\left(\int_{x_0}^{x+h}K(x,\alpha)f(\alpha)\,d\alpha-\int_{x_0}^{x}K(x,\alpha)f(\alpha)\,d\alpha\right)
\vspace{0.5em}\\&=&\displaystyle
\int_{x_0}^{x+h}\frac{K(x+h,\alpha)-K(x,\alpha)}h\,f(\alpha)\,d\alpha+\frac 1h\int_x^{x+h}K(x,\alpha)f(\alpha)\,d\alpha
\vspace{0.5em}\\&=&\displaystyle
\int_{x_0}^{x}\frac{K(x+h,\alpha)-K(x,\alpha)}h\,f(\alpha)\,d\alpha+\int_{x}^{x+h}\frac{K(x+h,\alpha)-K(x,\alpha)}h\,f(\alpha)\,d\alpha+\frac 1h\int_x^{x+h}K(x,\alpha)f(\alpha)\,d\alpha