$f^{\prime }(x)=\underset{h\to 0}{lim}\frac{(2(x+h{)}^6+\frac{3}{(x+h{)}^5}+\frac{\sqrt{(x+h)}}{h}-9)-(2x^5+\frac{3}{x^5}+\frac{\sqrt{x}}{4}-9)}{h}$

Step 1 ：First, we need to simplify the expression inside the limit. We can do this by expanding the terms and combining like terms.

Step 2 ：$=\underset{h\to 0}{lim}\frac{2(x^6+6x^{5}h+15x^4{h}^2+20x^3{h}^3+15x^2{h}^4+6x{h}^5+h^6)+\frac{3}{(x^5-5x^{4}h+10x^3{h}^2-10x^2{h}^3+5x{h}^4-h^5)}+\frac{\sqrt{x+h}}{h}-9-2x^5-\frac{3}{x^5}-\frac{\sqrt{x}}{4}+9}{h}$

Step 3 ：Next, we can cancel out the terms that are the same in the numerator.

Step 4 ：$=\underset{h\to 0}{lim}\frac{2(6x^{5}h+15x^4{h}^2+20x^3{h}^3+15x^2{h}^4+6x{h}^5+h^6)+\frac{3}{x^5}(1-\frac{5x^{4}h}{x^5}+\frac{10x^3{h}^2}{x^5}-\frac{10x^2{h}^3}{x^5}+\frac{5x{h}^4}{x^5}-\frac{h^5}{x^5})+\frac{\sqrt{x+h}}{h}-\frac{\sqrt{x}}{4}}{h}$

Step 5 ：Then, we can simplify the expression by dividing each term in the numerator by h.

Step 6 ：$=\underset{h\to 0}{lim}2(6x^5+15x^{4}h+20x^3{h}^2+15x^2{h}^3+6x{h}^4+h^5)+\frac{3}{x^5}(1-\frac{5x^4}{x}+\frac{10x^{3}h}{x^2}-\frac{10x^2{h}^2}{x^3}+\frac{5x{h}^3}{x^4}-\frac{h^4}{x^5})+\sqrt{x+h}-\frac{\sqrt{x}}{4h}$

Step 7 ：Now, we can take the limit as h approaches 0.

Step 8 ：$=2(6x^5)+\frac{3}{x^5}(1-\frac{5x^4}{x})+\sqrt{x}-\frac{\sqrt{x}}{4}\times 0$

Step 9 ：Finally, we simplify the expression to get the derivative of f.

Step 10 ：$=\boxed{{\displaystyle 12x^5-\frac{15}{x}+\sqrt{x}}}$