Question

find the derivative of:

2/x sqrt (9-x^2)

need step by step working out

Answer 1

Answer on Question #41084 – Math - Differential Calculus | Equations

First, function $h(x) = \sqrt{9 - x^2} = f\big(g(x)\big)$ is a composite function, where $f(x) = \sqrt{x}$, $g(x) = 9 - x^2$.

We know derivatives of the following functions:

$$(f(x))' = (\sqrt{x})' = \frac{1}{2\sqrt{x}},$$ $$(g(x))' = (9 - x^2)' = |derivative of a difference| = (9)' - (x^2)' = 0 - 2x = -2x$$

Derivative of a composite function is calculated by the formula

$$(\sqrt{9 - x^2})' = (h(x))' = (f(g(x))') = f'(g(x)) \times g'(x) = \frac{1}{2\sqrt{9 - x^2}} \times (-2x) = -\frac{x}{\sqrt{9 - x^2}}$$

Further, calculate

$$\begin{array}{l} \left(\frac{2}{x} \sqrt{9 - x^2}\right)' = |derivative of product| = \left(\frac{2}{x}\right)'(\sqrt{9 - x^2}) + \left(\frac{2}{x}\right)(\sqrt{9 - x^2})' = \\ = 2\left(\frac{1}{x}\right)'\sqrt{9 - x^2} + \frac{2}{x}(\sqrt{9 - x^2})' = 2 \times \frac{(-1)}{x^2} \times \sqrt{9 - x^2} + \frac{2}{x}(\sqrt{9 - x^2})' = \\ \end{array}$$

|use the previous result|

$$= -\frac{2\sqrt{9 - x^2}}{x^2} + \frac{2}{x} \times \frac{(-x)}{\sqrt{9 - x^2}} = -\frac{2\sqrt{9 - x^2}}{x^2} - \frac{2}{\sqrt{9 - x^2}} = \frac{-2}{x^2\sqrt{9 - x^2}} (9 - x^2 + x^2) = -\frac{18}{x^2\sqrt{9 - x^2}}.$$

Domain:

$$-3 < x < 0 \text{ or } 0 < x < 3$$

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