Answer to Question #150321 in Calculus for stefanus weyulu

Question

Let f(x) = π −
√
7x − 9. Use the definition of the derivative to find f
0
(x).

Answer 1

$f(x) = \pi - \sqrt{7x - 9} \\ f(x + \delta x) = \pi - \sqrt{7(x + \delta x) - 9} \\ f(x + \delta x) - f(x) = \pi - \sqrt{7(x + \delta x) - 9} - (\pi - \sqrt{7x - 9})\\ \begin{aligned} f(x + \delta x) - f(x) &= \sqrt{7x - 9} - \sqrt{7x - 9 + 7\delta x} \\&= -(7x - 9)^{\frac{1}{2}} - \frac{1}{2}(7x - 9)^{\frac{1}{2} - 1}(7\delta x) - \frac{\frac{1}{2}\left(\frac{1}{2} - 1\right)}{2!} (7x - 9)^{\frac{1}{2} - 2} (7\delta x)^2 - \\&\frac{\frac{1}{2}\left(\frac{1}{2} - 1\right)\left(\frac{1}{2} - 2\right)}{3!} (7x - 9)^{\frac{1}{2} - 3} (7\delta x)^3 - \cdots + (7x - 9)^{\frac{1}{2}} \\&= -(7x - 9)^{\frac{1}{2}} - \frac{(7x - 9)^{-\frac{1}{2}}}{2}(7\delta x) +\\& \frac{(7x - 9)^{-\frac{3}{2}}}{8}(7\delta x)^2 - \frac{(7x - 9)^{-\frac{5}{2}} (7\delta x)^3}{16} + \\&\cdots - (7x - 9)^{\frac{1}{2}} \\&= -\frac{(7x - 9)^{-\frac{1}{2}}}{2}(7\delta x) + \frac{(7x - 9)^{-\frac{3}{2}} (7\delta x)^2}{8} -\\& \frac{(7x - 9)^{-\frac{5}{2}} (7\delta x)^3}{16} + \cdots\\ \frac{f(x + \delta x) - f(x)}{\delta x} &=-\frac{7(7x - 9)^{-\frac{1}{2}}}{2} + \frac{(7x - 9)^{-\frac{3}{2}} (7\delta x)}{8} -\\& \frac{(7x - 9)^{-\frac{5}{2}} (7\delta x)^2}{16} + \cdots\\ \lim_{\delta x \to 0}\frac{f(x + \delta x) - f(x)}{\delta x} &= -\frac{7(7x - 9)^{-\frac{1}{2}}}{2} + \frac{(7x - 9)^{-\frac{3}{2}} (0)}{8} -\\& \frac{(7x - 9)^{-\frac{5}{2}} (0)^2}{16} + 0 + \cdots + 0 + \cdots\\ \therefore f'(x) &= \frac{-7}{2\sqrt{7x - 9}} \end{aligned}$