Derivative of $$$\frac{1}{t^{2} + x^{2}}$$$ with respect to $$$x$$$
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Find $$$\frac{d}{dx} \left(\frac{1}{t^{2} + x^{2}}\right)$$$.
Solution
The function $$$\frac{1}{t^{2} + x^{2}}$$$ is the composition $$$f{\left(g{\left(x \right)} \right)}$$$ of two functions $$$f{\left(u \right)} = \frac{1}{u}$$$ and $$$g{\left(x \right)} = t^{2} + x^{2}$$$.
Apply the chain rule $$$\frac{d}{dx} \left(f{\left(g{\left(x \right)} \right)}\right) = \frac{d}{du} \left(f{\left(u \right)}\right) \frac{d}{dx} \left(g{\left(x \right)}\right)$$$:
$${\color{red}\left(\frac{d}{dx} \left(\frac{1}{t^{2} + x^{2}}\right)\right)} = {\color{red}\left(\frac{d}{du} \left(\frac{1}{u}\right) \frac{d}{dx} \left(t^{2} + x^{2}\right)\right)}$$Apply the power rule $$$\frac{d}{du} \left(u^{n}\right) = n u^{n - 1}$$$ with $$$n = -1$$$:
$${\color{red}\left(\frac{d}{du} \left(\frac{1}{u}\right)\right)} \frac{d}{dx} \left(t^{2} + x^{2}\right) = {\color{red}\left(- \frac{1}{u^{2}}\right)} \frac{d}{dx} \left(t^{2} + x^{2}\right)$$Return to the old variable:
$$- \frac{\frac{d}{dx} \left(t^{2} + x^{2}\right)}{{\color{red}\left(u\right)}^{2}} = - \frac{\frac{d}{dx} \left(t^{2} + x^{2}\right)}{{\color{red}\left(t^{2} + x^{2}\right)}^{2}}$$The derivative of a sum/difference is the sum/difference of derivatives:
$$- \frac{{\color{red}\left(\frac{d}{dx} \left(t^{2} + x^{2}\right)\right)}}{\left(t^{2} + x^{2}\right)^{2}} = - \frac{{\color{red}\left(\frac{d}{dx} \left(t^{2}\right) + \frac{d}{dx} \left(x^{2}\right)\right)}}{\left(t^{2} + x^{2}\right)^{2}}$$The derivative of a constant is $$$0$$$:
$$- \frac{{\color{red}\left(\frac{d}{dx} \left(t^{2}\right)\right)} + \frac{d}{dx} \left(x^{2}\right)}{\left(t^{2} + x^{2}\right)^{2}} = - \frac{{\color{red}\left(0\right)} + \frac{d}{dx} \left(x^{2}\right)}{\left(t^{2} + x^{2}\right)^{2}}$$Apply the power rule $$$\frac{d}{dx} \left(x^{n}\right) = n x^{n - 1}$$$ with $$$n = 2$$$:
$$- \frac{{\color{red}\left(\frac{d}{dx} \left(x^{2}\right)\right)}}{\left(t^{2} + x^{2}\right)^{2}} = - \frac{{\color{red}\left(2 x\right)}}{\left(t^{2} + x^{2}\right)^{2}}$$Thus, $$$\frac{d}{dx} \left(\frac{1}{t^{2} + x^{2}}\right) = - \frac{2 x}{\left(t^{2} + x^{2}\right)^{2}}$$$.
Answer
$$$\frac{d}{dx} \left(\frac{1}{t^{2} + x^{2}}\right) = - \frac{2 x}{\left(t^{2} + x^{2}\right)^{2}}$$$A