ANSWER
[tex]f^{\prime}(x)=\frac{\sqrt{x}(2x-9)}{2(x-3)^{3/2}}[/tex]EXPLANATION
Before finding the derivative of this function, we can simplify it. This will also simplify the process of derivation.
We include x inside the square root if we square it, and the same for the denominator (x - 3),
[tex]f(x)=\frac{x\sqrt{x(x-3)}}{x-3}=\sqrt{\frac{x^2\cdot x(x-3)}{(x-3)^2}}=\sqrt{\frac{x^3}{x-3}}[/tex]Now we apply the chain rule,
[tex]h^{\prime}(g(x))=h^{\prime}(g)\cdot g^{\prime}(x)[/tex]In this case, h(g) is the function f and g(x) is the expression inside the square root,
[tex]f(g)=\sqrt{g};g(x)=\frac{x^3}{x-3}[/tex]Let's find the derivative of f(g). Remember that the square root can be written as a power with exponent 1/2,
[tex]f(g)=g^{1/2}\Rightarrow f^{\prime}(g)=\frac{1}{2}g^{1/2-1}=\frac{1}{2}g^{-1/2}=\frac{1}{2\sqrt{g}}[/tex]Then, find the derivative of g(x). To do so, we have to apply the quotient rule,
[tex]\left(\frac{u(x)}{v(x)}\right)^{\prime}=\frac{u^{\prime}(x)\cdot v(x)-u(x)\cdot v^{\prime}(x)}{v^2(x)}[/tex]In this case, u(x) = x³ and v(x) = x - 3. Let's find the derivative of each and the square of v(x),
[tex]u^{\prime}(x)=3x^{3-1}=3x^2[/tex][tex]v^{\prime}(x)=x^{1-1}-0=1[/tex][tex]v^2(x)=(x-3)^2[/tex]So, the derivative of g(x) is,
[tex]g^{\prime}(x)=\frac{3x^2(x-3)-x^3}{(x-3)^2}=\frac{3x^3-9x^2-x^3}{(x-3)^2}=\frac{2x^3-9x^2}{(x-3)^2}[/tex]Finally, plug in g'(x) and f'(g(x)) in the chain rule expression we found above,
[tex]f^{\prime}(x)=\frac{1}{2\sqrt{\frac{x^3}{x-3}}}\cdot\frac{2x^3-9x^2}{(x-3)^2}[/tex]This expression can be simplified. First, distribute the square root of the first factor and rewrite it using fractional exponents,
[tex]f^{\prime}(x)=\frac{(x-3)^{1/2}}{2x^{3/2}}\cdot\frac{2x^{3}-9x^{2}}{(x-3)^{2}}[/tex]Simplify the exponents of the factor (x - 3),
[tex]f^{\prime}(x)=\frac{(x-3)^{1/2-2}}{2x^{3/2}}\cdot2x^3-9x^2=\frac{(x-3)^{-3/2}}{2x^{3/2}}\cdot2x^3-9x^2=\frac{2x^3-9x^2}{2x^{3/2}(x-3)^{3/2}}[/tex]We can also take x² as a common factor in the numerator to simplify that with the denominator,
[tex]f^{\prime}(x)=\frac{x^2(2x^-9)}{2x^{3/2}(x-3)^{3/2}}=\frac{x^{2-3/2}(2x^-9)}{2(x-3)^{3/2}}=\frac{x^{1/2}(2x-9)}{2(x-3)^{3/2}}[/tex]Hence, the derivative of the function is,
[tex]f^{\prime}(x)=\frac{\sqrt{x}(2x-9)}{2(x-3)^{3/2}}[/tex]