If \(f(x)=x\sqrt{2x-3}\), then \(f'(x)=\)

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Follow the product rule.

$$ \frac{d(uv)}{dx}=u\frac{dv}{dx}+v\frac{du}{dx} $$

In our case, \(u=x\) and \(v=\sqrt{2x-3}\).

Substituting the values into the product rule:

$$ f'(x)=\frac{d}{dx}(x\sqrt{2x-3})=x\cdot\frac{d}{dx}\sqrt{2x-3}+\sqrt{2x-3}\cdot\frac{d}{dx}x $$ $$ f'(x)= x\left(\frac{1}{\sqrt{2x-3}}\right) + \sqrt{2x-3}(1) $$ $$ f'(x)= \frac{x}{\sqrt{2x-3}}+\sqrt{2x-3} $$ $$ f'(x)= \frac{x}{\sqrt{2x-3}}+\frac{2x-3}{\sqrt{2x-3} }$$ $$ f'(x)=\frac{x+2x-3}{\sqrt{2x-3}} $$ $$ = \boxed{\frac{3x-3}{\sqrt{2x-3}}}$$