By the definition of $F$,

$$\frac{\partial}{\partial \lambda}F({q}_{0}+\lambda q)=\frac{\partial}{\partial \lambda}{\int}_{D}L(x,{q}_{0}+\lambda q,d{q}_{0}+\lambda dq){d}^{m}x$$ 


$$=\frac{\partial}{\partial \lambda}\left({\int}_{x{x}_{0}\le r}L(x,{q}_{0}+\lambda q,d{q}_{0}+\lambda dq){d}^{m}x+{\int}_{\genfrac{}{}{0pt}{}{x\in D}{x{x}_{0}>r}}L(x,{q}_{0}+\lambda q,d{q}_{0}+\lambda dq){d}^{m}x\right)$$ 

By the condition imposed on $q$, the derivative of the second integral is zero. Since the integrand of the first integral and its first derivatives^{} are continuous^{} and the closed ball is compact, the integrand and its first derivatives are uniformly continuous, so it is permissible to interchange differentiation and integration. Hence,

$$\frac{\partial}{\partial \lambda}F({q}_{0}+\lambda q)={\int}_{x{x}_{0}\le r}\frac{\partial L(x,{q}_{0}+\lambda q,d{q}_{0}+\lambda dq)}{\partial \lambda}{d}^{m}x$$ 
