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Question Number 169358 by Mastermind last updated on 29/Apr/22

$$Differentiatefromfirstprinciple$$$$y=\frac{1}{x^2+5}$$

Answered by bobhans last updated on 29/Apr/22

$$\frac{dy}{dx}=\underset{p\to 0}{\lim }\frac{f(x+p)-f\left(x\right)}{p}$$$$=\underset{p\to 0}{\lim }\frac{\frac{1}{{(x+p)}^2+5}-\frac{1}{x^2+5}}{p}$$$$=\underset{p\to 0}{\lim }\frac{(x^2+5)-(x^2+2px+p^2+5)}{p(x^2+5)[{(x+p)}^2+5]}$$$$=\underset{p\to 0}{\lim }\frac{1}{(x^2+5)[{(x+p)}^2+5]}.\underset{p\to 0}{\lim }\frac{-2px-p^2}{p}$$$$=\frac{1}{{(x^2+5)}^2}.\underset{p\to 0}{\lim }\frac{-2x-p}{1}=\frac{-2x}{{(x^2+5)}^2}$$

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