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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.

Nick Dong
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2 Answers2

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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.

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  • 27,005
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We have, $$G_0’(x)= \frac{F_0(u)F_0’(u+(x-1)F_1(u))F_1(u)-0}{F_0(u)^2} =\frac{F_1(u)}{F_0(u)}[F_0’(u+(x-1)F_1(u))]$$

Thus, $$G_0’(1)=\frac{F_1(u)}{F_0(u)}[F_0’(u)]$$

Thus, $$G_1(x)=\frac{F_0’(u+(x-1)F_1(u))}{F_0’(u)}$$ Multiplying and dividing by $F_0’(1)$ gives us, $$\boxed{G_1(x)=\frac{F_1(u+(x-1)F_1(u))}{F_1(u)}}$$

  • Seems right. I was confused with $F_0(u)$. I thought that it should be treated as constant, which should not be considered during caculation. – Nick Dong Dec 10 '17 at 12:59
  • Thanks. That is helpful. But @TheSimpliFire 's steps is more close to my expectation. – Nick Dong Dec 10 '17 at 13:36