Find `(dy)/(dx)` if, `y=((x^2-1)^3(2x-1))/(sqrt((x-3)(4x-1)))`

Explanation

`text { Let } y=frac{(x^{2}-1)^{3}(2 x-1)}{sqrt{(x-3)(4 x-1)}} `

`Rightarrow y=frac{(x^{2}-1)^{3}(2 x-1)}{((x-3)(4 x-1))^{frac{1}{2}}} `

`Rightarrow y=frac{(x^{2}-1)^{3}(2 x-1)}{(x-3)^{frac{1}{2}}(4 x-1)^{frac{1}{2}}}`

Take log both sides:

`Rightarrow log y= log {frac{(x^{2}-1)^{3}(2 x-1)}{(x-3)^{frac{1}{2}}(4 x-1)^{frac{1}{2}}}} `

`[log (a b).=log a+log b ; log (frac{a}{b})=log a-log b] `

`Rightarrow log y={log (x^{2}-1)^{3}+log (2 x-1)} `

`-{log (x-3)^{frac{1}{2}}+log (4 x-1)^{frac{1}{2}}} `

`Rightarrow log y= log (x^{2}-1)^{3}+log (2 x-1) `

`- log (x-3)^{frac{1}{2}}-log (4 x-1)^{frac{1}{2}}`

`Rightarrow log y= 3 log (x^{2}-1)+log (2 x-1)`

`-frac{1}{2} log (x-3)-frac{1}{2} log (4 x-1) `

`{log x^{a}=a log x}`

DIfferentlating wIth respect to `{X}` :

`Rightarrow frac{d(log y)}{d x}=3 frac{d(log (x^{2}-1))}{d x}+frac{d(log (2 x-1))}{d x}-frac{1}{2} frac{d(log (x-3))}{d x}`

`-frac{1}{2} frac{d(log (4 x-1))}{d x}`

`{` Using chain rule, `frac{d(u+a)}{d x}=frac{d u}{d x}+frac{d a}{d {x}}` where a and u are any variables `}`

`Rightarrow frac{1}{y} frac{d y}{d x}=frac{3}{x^{2}-1} frac{d(x^{2}-1)}{d x}+frac{1}{(2 x-1)} frac{d(2 x-1)}{d x}`

`-frac{1}{2(x-3)} frac{d(x-3)}{d x}-frac{1}{2(4 x-1)} frac{d(4 x-1)}{d x}` Using chain rule, `frac{d(a+u)}{d x}=frac{d u}{d x}` where `a` is any constant and `u` is any variable; `}`

`Rightarrow frac{d(u^{n})}{d x}=n u^{n-1} frac{d u}{d x}`

`Rightarrow frac{1}{d x}=frac{d y}{d x}=frac{6 x}{x^{2}-1}+frac{2}{(2 x-1)}-frac{1}{2(x-3)}-frac{2}{2(4 x-1)}`

`Rightarrow frac{d y}{d x}=y{frac{6 x}{x^{2}-1}+frac{2}{(2 x-1)}-frac{1}{2(x-3)}-frac{2}{(4 x-1)}-frac{2}{(4 x-1)}}`