Answer:
[tex]f^{\prime}\left(x\right)\ =\ -\frac{65}{2}x^{\frac{11}{2}}\ +\frac{49}{2}x^{-\frac{9}{2}}[/tex]
or
[tex]f^{\prime}\left(x\right)\ =\ -32.5x^{5.5}\ +\ 24.5x^{-4.5}[/tex]
Step-by-step explanation:
Rather than solving this question in a more complex method by directly using the product rule and quotient rule, it can first be considered to perform some algebraic manipulation (index laws) to simplify the expression before taking the derivative.
[tex]\begin{large}\begin{array}{l}f\left(x\right)\ =\ -5x^6\ \sqrt{x}\ +\ \frac{-7}{x^3\ \sqrt{x}}\\\\f\left(x\right)\ =\ -5x^6\cdot x^{\frac{1}{2}}\ +\ \frac{-7}{x^3\cdot x^{\frac{1}{2}}}\\\\f\left(x\right)\ =\ -5x^{6\ +\ \frac{1}{2}}\ +\ \frac{-7}{x^{3\ +\ \frac{1}{2}}}\\\\f\left(x\right)\ =\ -5x^{\frac{13}{2}}\ +\ \frac{-7}{x^{\frac{7}{2}}}\\\\f\left(x\right)\ =\ -5x^{\frac{13}{2}}\ -7x^{-\frac{7}{2}}\end{array}[/tex]
Now, the derivative of the function can be calculated simply by only using the power rule, which yields
[tex]\begin{large}\begin{array}{l}f\left(x\right)\ =\ -5x^{\frac{13}{2}}\ -7x^{-\frac{7}{2}}\\\\f^{\prime}\left(x\right)\ =\ \left(-5\right)\left(\frac{13}{2}\right)\left(x^{\frac{13}{2}\ -\ 1}\right)\ -\ \left(7\right)\left(-\frac{7}{2}\right)\left(x^{-\frac{7}{2}\ -\ 1}\right)\\\\f^{\prime}\left(x\right)\ =\ -\frac{65}{2}x^{\frac{11}{2}}\ +\frac{49}{2}x^{-\frac{9}{2}}\\\\f^{\prime}\left(x\right)\ =\ -32.5x^{5.5}\ +\ 24.5x^{-4.5}\end{array}\\\end{large}[/tex]
