Given:
[tex]y=\dfrac{x}{\sqrt{x-1}}[/tex]
To find:
The derivative of the given problem.
Solution:
Chain rule : [tex]\dfrac{d}{dx}f(g(x))=f'(g(x))\cdot g'(x)[/tex]
Quotient rule : [tex]\dfrac{d}{dx}\left(\dfrac{f(x)}{g(x)}\right)=\dfrac{g(x)f'(x)-f(x)g'(x)}{[g(x)]^2}[/tex]
We have,
[tex]y=\dfrac{x}{\sqrt{x-1}}[/tex]
Differentiating with respect to x using chain rule and quotient rule, we get
[tex]\dfrac{dy}{dx}=\dfrac{\sqrt{x-1}\dfrac{d}{dx}(x)-x\dfrac{d}{dx}(\sqrt{x-1})}{(\sqrt{x-1})^2}[/tex]
[tex]\dfrac{dy}{dx}=\dfrac{\sqrt{x-1}(1)-x\dfrac{1}{2\sqrt{x-1}}\dfrac{d}{dx}(x-1)}{(x-1)}[/tex]
[tex]\dfrac{dy}{dx}=\dfrac{\sqrt{x-1}-\dfrac{x}{2\sqrt{x-1}}(1-0)}{x-1}[/tex]
[tex]\dfrac{dy}{dx}=\dfrac{\sqrt{x-1}-\dfrac{x}{2\sqrt{x-1}}}{x-1}[/tex]
[tex]\dfrac{dy}{dx}=\dfrac{\dfrac{2x-2-x}{2(x-1)^{\frac{1}{2}}}}{(x-1)}[/tex]
[tex]\dfrac{dy}{dx}=\dfrac{x-2}{2(x-1)^{\frac{3}{2}}}[/tex]
Therefore, the required derivative is [tex]\dfrac{dy}{dx}=\dfrac{x-2}{2(x-1)^{\frac{3}{2}}}[/tex].