Chapter 11.3, Problem 39E

To determine

To find: the derivative of the function.

Answer to Problem 39E

The derivative of the function is $-\frac{1}{2{\left(x-9\right)}^{3/2}}$

Explanation of Solution

Given:

  $f\left(x\right)=\frac{1}{\sqrt{x-9}}$

Calculation:

Remember that the derivative of the function f is given by

  $f^{\prime }\left(x\right)=\underset{h\to 0}{\lim }\frac{f\left(x+h\right)-f\left(x\right)}{h}$

Therefore, to find the derivative of this function, substitute into the given function, expand the terms, divide out, and simplify

  $\begin{array}{c}{f}^{\prime }\left(x\right)=\underset{h\to 0}{\lim }\frac{\frac{1}{\sqrt{x+h-9}}-\frac{1}{\sqrt{x-9}}}{h}\\ =\underset{h\to 0}{\lim }\frac{\frac{1}{\sqrt{x+h-9}}-\frac{1}{\sqrt{x-9}}}{h}*\frac{\sqrt{x+h-9}\sqrt{x-9}}{\sqrt{x+h-9}\sqrt{x-9}}\\ =\underset{h\to 0}{\lim }\frac{\sqrt{x-9}-\sqrt{x+h-9}}{h\sqrt{x+h-9}\sqrt{x-9}}\\ =\underset{h\to 0}{\lim }\frac{\sqrt{x-9}-\sqrt{x+h-9}}{h\sqrt{x+h-9}\sqrt{x-9}}*\frac{\sqrt{x-9}+\sqrt{x+h-9}}{\sqrt{x-9}+\sqrt{x+h-9}}\\ =\underset{h\to 0}{\lim }\frac{x-9-\left(x+h-9\right)}{h\sqrt{x+h-9}\sqrt{x-9}\left(\sqrt{x-9}+\sqrt{x+h-9}\right)}\\ =\underset{h\to 0}{\lim }\frac{-h}{h\sqrt{x+h-9}\sqrt{x-9}\left(\sqrt{x-9}+\sqrt{x+h-9}\right)}\\ =\underset{h\to 0}{\lim }-\frac{1}{\sqrt{x+h-9}\sqrt{x-9}\left(\sqrt{x-9}+\sqrt{x+h-9}\right)}\\ =-\frac{1}{\left(x-9\right)\left(2\sqrt{x-9}\right)}\left(\text{By direct substitution}\text{.}\right)\\ =-\frac{1}{2{\left(x-9\right)}^{3/2}}\end{array}$

Conclusion:

The derivative of the function is $-\frac{1}{2{\left(x-9\right)}^{3/2}}$