Question

find dy/dx if y = (1-2x)^2/3 (1 + 3x)^-3/4
(1 - 6x)^5/6 (1 + 7x)^-6/7​

Answer 1

Answer:

Please find the attachment

Answer 2

Given:

$y=\frac{(1-2 x)^{2 / 3}(1+3 x)^{-3 / 4}}{(1-6 x)^{5 / 6}(1+7 x)^{-6 / 7}}$

$\text { Apply log on both sides. }$

$\log y=\log \left(\frac{(1-2 x)^{2 / 3}(1+3 x)^{-3 / 4}}{(1-6 x)^{5 / 6}(1+7 x)^{-6 / 7}}\right)$  $\log \frac{m}{n}=\log m-\log n$

$\left.\log y=\log \left[(1-2 x)^{1 / 3}(1+3 x)^{-3 / 4}\right)-\log (1-6 x)^{5 / 6}(1+7 x)^{-6 / 7}\right)$

$=\log (1-2 x)^{2 / 3}+\log (1+3 x)^{-3 / 4}-\log (1-6 x)^{5 / 6}-\log (1+7 x)^{-6 / 7}$

$\log y=\frac{2}{3} \log (1-2 x)+\left(\frac{-3}{4}\right) \log (1+3 x)-\frac{5}{6} \log (1-6 x)+\frac{6}{7} \log (1+7 x)$

$\text { Differentiate w.r.t } x$

$\frac{d}{d x}(\log y)=\frac{d}{d x}\left(\frac{2}{3} \log (1-2 x)-\frac{3}{4} \log (1+3 x)+\frac{5}{6} \log (1-6 x)+\frac{6}{7} \log (1+7 x)\right.)$

$\frac{d}{d x} \log x=\frac{1}{x}$

$\frac{1}{y} \frac{d y}{d x}=\frac{2}{3} \ \frac{d}{d x}(1-2 x) \frac{1}{1-2 x}-\frac{3}{4} \frac{1}{1+3 x} \frac{d}{d x}(1+3 x)-\frac{3}{6} \frac{1}{(1-6 x)} \frac{d x}{d x}(1-6 x)$

$\left.+\frac{6}{7} \frac{1}{1+7 x} \frac{d}{d x}(1+7 x)\right]$

$\frac{d y}{d x}=y\left[\frac{2(-2)}{3 (1-2 x)}-\frac{3(3)}{4(1+3 x)}+\frac{5(-6)}{6(1-6 x)}+\frac{6(7)}{7(1+7 x)}\right]$

$\left.\frac{d y}{d x}=y[\frac{-4}{3(1-2x)} -\frac{9}{4(1+3x)}+\frac{5}{1-6 x}+\frac{6}{1+7 x}\right]$

$\left.\frac{d y}{d x}=\frac{(1-2 x)^{2 / 3}(1+7 x)^{-3 / 4}}{(1-6 x)^{5 / 6}(1+2 x)^{-1 / 7}} [ \frac{3}{1-6 x}+\frac{6}{1+7 x}-\frac{4}{3(1-2 x)}-\frac{9}{4(1+3 x)}\right]$