Show that $G:[c,d]\to \mathbb R$ defined by $G(x)=\int_0^{f(x)}F(t)dt, x\in [c,d] $ is differentiable and $G'(x)=F\circ f(x) \cdot f'(x).$

Question

Let $F$ be continuous on $[a,b].$ let $f:[c,d]\to \mathbb R$ be differential function satisfy $f([c,d])\subseteq [a,b].$ Show that $G:[c,d]\to \mathbb R$ defined by $G(x)=\int_0^{f(x)}F(t)dt, x\in [c,d] $ is differentiable and $G'(x)=F\circ f(x) \cdot f'(x).$

My attempt

$\lim_{h\to 0} \frac{ G(x+h)-G(x)}{h}=\lim_{h\to 0} \frac{ \int_0^{f(x+h)}F(t)dt-\int_0^{f(x)}F(t)dt}{h}=\lim_{h\to 0} \frac{ \int_{f(x)}^{f(x+h)}F(t)dt}{h}$ I am not able to go beyond this.