Derivative of $$$\frac{1}{\sqrt{x^{2} - 3 x + 9}}$$$

The calculator will find the derivative of $$$\frac{1}{\sqrt{x^{2} - 3 x + 9}}$$$, with steps shown.

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Find $$$\frac{d}{dx} \left(\frac{1}{\sqrt{x^{2} - 3 x + 9}}\right)$$$.

Solution

The function $$$\frac{1}{\sqrt{x^{2} - 3 x + 9}}$$$ is the composition $$$f{\left(g{\left(x \right)} \right)}$$$ of two functions $$$f{\left(u \right)} = \frac{1}{\sqrt{u}}$$$ and $$$g{\left(x \right)} = x^{2} - 3 x + 9$$$.

Apply the chain rule $$$\frac{d}{dx} \left(f{\left(g{\left(x \right)} \right)}\right) = \frac{d}{du} \left(f{\left(u \right)}\right) \frac{d}{dx} \left(g{\left(x \right)}\right)$$$:

$${\color{red}\left(\frac{d}{dx} \left(\frac{1}{\sqrt{x^{2} - 3 x + 9}}\right)\right)} = {\color{red}\left(\frac{d}{du} \left(\frac{1}{\sqrt{u}}\right) \frac{d}{dx} \left(x^{2} - 3 x + 9\right)\right)}$$

Apply the power rule $$$\frac{d}{du} \left(u^{n}\right) = n u^{n - 1}$$$ with $$$n = - \frac{1}{2}$$$:

$${\color{red}\left(\frac{d}{du} \left(\frac{1}{\sqrt{u}}\right)\right)} \frac{d}{dx} \left(x^{2} - 3 x + 9\right) = {\color{red}\left(- \frac{1}{2 u^{\frac{3}{2}}}\right)} \frac{d}{dx} \left(x^{2} - 3 x + 9\right)$$

Return to the old variable:

$$- \frac{\frac{d}{dx} \left(x^{2} - 3 x + 9\right)}{2 {\color{red}\left(u\right)}^{\frac{3}{2}}} = - \frac{\frac{d}{dx} \left(x^{2} - 3 x + 9\right)}{2 {\color{red}\left(x^{2} - 3 x + 9\right)}^{\frac{3}{2}}}$$

The derivative of a sum/difference is the sum/difference of derivatives:

$$- \frac{{\color{red}\left(\frac{d}{dx} \left(x^{2} - 3 x + 9\right)\right)}}{2 \left(x^{2} - 3 x + 9\right)^{\frac{3}{2}}} = - \frac{{\color{red}\left(\frac{d}{dx} \left(x^{2}\right) - \frac{d}{dx} \left(3 x\right) + \frac{d}{dx} \left(9\right)\right)}}{2 \left(x^{2} - 3 x + 9\right)^{\frac{3}{2}}}$$

Apply the power rule $$$\frac{d}{dx} \left(x^{n}\right) = n x^{n - 1}$$$ with $$$n = 2$$$:

$$- \frac{{\color{red}\left(\frac{d}{dx} \left(x^{2}\right)\right)} + \frac{d}{dx} \left(9\right) - \frac{d}{dx} \left(3 x\right)}{2 \left(x^{2} - 3 x + 9\right)^{\frac{3}{2}}} = - \frac{{\color{red}\left(2 x\right)} + \frac{d}{dx} \left(9\right) - \frac{d}{dx} \left(3 x\right)}{2 \left(x^{2} - 3 x + 9\right)^{\frac{3}{2}}}$$

The derivative of a constant is $$$0$$$:

$$- \frac{2 x + {\color{red}\left(\frac{d}{dx} \left(9\right)\right)} - \frac{d}{dx} \left(3 x\right)}{2 \left(x^{2} - 3 x + 9\right)^{\frac{3}{2}}} = - \frac{2 x + {\color{red}\left(0\right)} - \frac{d}{dx} \left(3 x\right)}{2 \left(x^{2} - 3 x + 9\right)^{\frac{3}{2}}}$$

Apply the constant multiple rule $$$\frac{d}{dx} \left(c f{\left(x \right)}\right) = c \frac{d}{dx} \left(f{\left(x \right)}\right)$$$ with $$$c = 3$$$ and $$$f{\left(x \right)} = x$$$:

$$- \frac{2 x - {\color{red}\left(\frac{d}{dx} \left(3 x\right)\right)}}{2 \left(x^{2} - 3 x + 9\right)^{\frac{3}{2}}} = - \frac{2 x - {\color{red}\left(3 \frac{d}{dx} \left(x\right)\right)}}{2 \left(x^{2} - 3 x + 9\right)^{\frac{3}{2}}}$$

Apply the power rule $$$\frac{d}{dx} \left(x^{n}\right) = n x^{n - 1}$$$ with $$$n = 1$$$, in other words, $$$\frac{d}{dx} \left(x\right) = 1$$$:

$$- \frac{2 x - 3 {\color{red}\left(\frac{d}{dx} \left(x\right)\right)}}{2 \left(x^{2} - 3 x + 9\right)^{\frac{3}{2}}} = - \frac{2 x - 3 {\color{red}\left(1\right)}}{2 \left(x^{2} - 3 x + 9\right)^{\frac{3}{2}}}$$

Simplify:

$$- \frac{2 x - 3}{2 \left(x^{2} - 3 x + 9\right)^{\frac{3}{2}}} = \frac{\frac{3}{2} - x}{\left(x^{2} - 3 x + 9\right)^{\frac{3}{2}}}$$

Thus, $$$\frac{d}{dx} \left(\frac{1}{\sqrt{x^{2} - 3 x + 9}}\right) = \frac{\frac{3}{2} - x}{\left(x^{2} - 3 x + 9\right)^{\frac{3}{2}}}$$$.

Answer

$$$\frac{d}{dx} \left(\frac{1}{\sqrt{x^{2} - 3 x + 9}}\right) = \frac{\frac{3}{2} - x}{\left(x^{2} - 3 x + 9\right)^{\frac{3}{2}}}$$$A


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