Differentiate-from-first-principle-y-1-x-2-5-

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

D i f f e r e n t i a t e f r o m f i r s t p r i n c i p l e

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

Answered by bobhans last updated on 29/Apr/22

\frac{d y}{d x} = \lim_{p \to 0} \frac{f (x + p) - f (x)}{p}

= \lim_{p \to 0} \frac{\frac{1}{{(x + p)}^{2} + 5} - \frac{1}{x^{2} + 5}}{p}

= \lim_{p \to 0} \frac{(x^{2} + 5) - (x^{2} + 2 p x + p^{2} + 5)}{p (x^{2} + 5) [{(x + p)}^{2} + 5]}

= \lim_{p \to 0} \frac{1}{(x^{2} + 5) [{(x + p)}^{2} + 5]} . \lim_{p \to 0} \frac{- 2 p x - p^{2}}{p}

= \frac{1}{{(x^{2} + 5)}^{2}} . \lim_{p \to 0} \frac{- 2 x - p}{1} = \frac{- 2 x}{{(x^{2} + 5)}^{2}}

Facebook Tweet Pin