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

Summary Submit Skip Question

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:

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