Question

If $y=\frac{ax+b}{cx+d}$, prove that 2y₁y₃ = 3(y₂)².

Answer 1

Here,
$$\begin{gathered} y=\frac{ax+b}{cx+d} \\ \mathrm{Differentiating}\mathrm{w}.\mathrm{r}.\mathrm{t}.x\mathit{,}\mathit{}\mathrm{we}\mathrm{get} \\ {y}_1=\frac{acx+ad-acx-bc}{{\left(cx+d\right)}^2} \\ =\frac{ad-bc}{{\left(cx+d\right)}^2} \\ \mathrm{Differentiating}\mathrm{again}\mathrm{w}.\mathrm{r}.\mathrm{t}.x\mathit{,}\mathit{}\mathrm{we}\mathrm{get} \\ {y}_2=\frac{-2(ad-bc)c}{{\left(cx+d\right)}^3} \\ \mathrm{Differentiating}\mathrm{again}\mathrm{w}.\mathrm{r}.\mathrm{t}.x\mathit{,}\mathit{}\mathrm{we}\mathrm{get} \\ {y}_3=\frac{6(ad-bc)c^2}{{\left(cx+d\right)}^4} \\ \Rightarrow y_3=\frac{6(ad-bc)c^2}{{\left(cx+d\right)}^4}\times \frac{{\left(cx+d\right)}^2}{{\left(cx+d\right)}^2}\left(\frac{ad-bc}{ad-bc}\right) \\ \Rightarrow y_3=3{\left\{\frac{-2\left(ad-bc\right)c}{{\left(cx+d\right)}^3}\right\}}^2\times \frac{1}{2y_1} \\ \Rightarrow 2y_1{y}_3=3{\left(y_2\right)}^2 \end{gathered}$$

Hence proved.