Question

Find $\frac{dy}{dx}$

$y=\frac{{\left(x^2-1\right)}^3\left(2x-1\right)}{\sqrt{\left(x-3\right)\left(4x-1\right)}}$

Answer 1

$$\begin{gathered} \mathrm{We}\mathrm{have},y=\frac{{\left(x^2-1\right)}^3\left(2x-1\right)}{\sqrt{\left(x-3\right)\left(4x-1\right)}}...\left(i\right) \\ \Rightarrow y=\frac{{\left(x^2-1\right)}^3\left(2x-1\right)}{{\left(x-3\right)}^{{\displaystyle \frac{1}{2}}}{\left(4x-1\right)}^{{\displaystyle \frac{1}{2}}}} \end{gathered}$$
Taking log on both sides,
$$\begin{gathered} \log y=\log \left[\frac{{\left(x^2-1\right)}^3\left(2x-1\right)}{{\left(x-3\right)}^{{\displaystyle \frac{1}{2}}}{\left(4x-1\right)}^{{\displaystyle \frac{1}{2}}}}\right] \\ \Rightarrow \log y=\log {\left(x^2-1\right)}^3+\log \left(2x-1\right)-\log {\left(x-3\right)}^{\frac{1}{2}}-\log {\left(4x-1\right)}^{\frac{1}{2}} \\ \Rightarrow \log y=3\log \left(x^2-1\right)+\log \left(2x-1\right)-\frac{1}{2}\log \left(x-3\right)-\frac{1}{2}\log \left(4x-1\right) \end{gathered}$$
Differentiating with respect to x using chain rule,
$$\begin{gathered} \frac{1}{y}\frac{dy}{dx}=3\frac{d}{dx}\left\{\log \left(x^2-1\right)\right\}+\frac{d}{dx}\left\{\log \left(2x-1\right)\right\}-\frac{1}{2}\frac{d}{dx}\left\{\log \left(x-3\right)\right\}-\frac{1}{2}\left\{\log \left(4x-1\right)\right\} \\ \Rightarrow \frac{1}{y}\frac{dy}{dx}=3\left(\frac{1}{x^2-1}\right)\frac{d}{dx}\left(x^2-1\right)+\frac{1}{\left(2x-1\right)}\frac{d}{dx}\left(2x-1\right)-\frac{1}{2}\left(\frac{1}{x-3}\right)\frac{d}{dx}\left(x-3\right)-\frac{1}{2}\frac{1}{\left(4x-1\right)}\frac{d}{dx}\left(4x-1\right) \\ \Rightarrow \frac{1}{y}\frac{dy}{dx}=3\left(\frac{1}{x^2-1}\right)\left(2x\right)+\frac{1}{2x-1}\left(2\right)-\frac{1}{2}\left(\frac{1}{x-3}\right)\left(1\right)-\frac{1}{2}\left(\frac{1}{4x-1}\right)\left(4\right) \\ \Rightarrow \frac{1}{y}\frac{dy}{dx}=\left[\frac{6x}{x^2-1}+\frac{2}{2x-1}-\frac{1}{2\left(x-3\right)}-\frac{2}{4x-1}\right] \\ \Rightarrow \frac{dy}{dx}=y\left[\frac{6x}{x^2-1}+\frac{2}{2x-1}-\frac{1}{2\left(x-3\right)}-\frac{2}{4x-1}\right] \\ \Rightarrow \frac{dy}{dx}=\frac{{\left(x^2-1\right)}^3\left(2x-1\right)}{\sqrt{\left(x-3\right)\left(4x-1\right)}}\left[\frac{6x}{x^2-1}+\frac{2}{2x-1}-\frac{1}{2\left(x-3\right)}-\frac{2}{4x-1}\right]\left[\mathrm{using}\mathrm{equation}\left(\mathrm{i}\right)\right] \end{gathered}$$