derivative caculation I met a problem that described as follow. I am not sure the title is suitable for my problem. If you have any advices about the title or the description, please comment below.
Aready known:
$$F_1(x)=\frac{F_0^\prime(x)}{F_0^\prime(1)}$$
$$G_0(x)=\frac{1}{F_0(u)}F_0(u+(x-1)F_1(u))$$
$$G_1(x)=\frac{G_0^\prime(x)}{G_0^\prime(1)}$$
Expected result that I need to obtain:
$$G_1(x)=\frac{G_0^\prime(x)}{G_0^\prime(1)}=\frac{1}{F_1(u)}F_1(u+(x-1)F_1(u))$$
I tried:
$$G_0^\prime(x)=\frac{\mathrm{d}G_0(x)}{\mathrm{d}x}=\frac{\mathrm{d}(\frac{1}{F_0(u)}F_0(u+(x-1)F_1(u)))}{\mathrm{d}x}
=\frac{F_1(u)}{F_0(u)}\frac{\mathrm{d}(F_0(u+(x-1)F_1(u)))}{\mathrm{d}x}$$
$$G_0^\prime(1)=\left . G_0^\prime(x)\right\vert_{x=1}=\frac{F_1(u)}{F_0(u)}\left . \frac{\mathrm{d}F_0(u+(x-1)F_1(u))}{\mathrm{d}x}\right\vert_{x=1}$$
$$F_1(u+(x-1)F_1(u))=\frac{\frac{\mathrm{d}(F_0(u+(x-1)F_1(u)))}{\mathrm{d}x}}{\left . \frac{\mathrm{d}(F_0(u+(x-1)F_1(u)))}{\mathrm{d}x}\right\vert_{x=1}}$$
But I did not get the expected result. P.S. $u$ is not a function of $x$. So $u$ should be treated as a constant. 
Could anyone do me this favor?  Please do not mislead by the step that I tried.

We have $$G_0'(x)=\frac{F_1(u)}{F_0(u)}F_0'(u+(x-1)F_1(u))$$ and substituing $x=1$ gives $$G_0'(1)=\frac{F_1(u)}{F_0(u)}F_0'(u+(1-1)F_1(u))=\frac{F_1(u)}{F_0(u)}F_0'(u)$$ Hence $$G_1(x)=\dfrac{\dfrac{F_1(u)}{F_0(u)}F_0'(u+(x-1)F_1(u))}{\dfrac{F_1(u)}{F_0(u)}F_0'(u)}=\dfrac{\dfrac{F_0'(u+(x-1)F_1(u))}{F_0'(1)}}{\dfrac{F_0'(u)}{F_0'(1)}}=\dfrac{F_1(u+(x-1)F_1(u))}{F_1(u)}$$ as required.
