Question Number 169358 by Mastermind last updated on 29/Apr/22
$${Differentiate}\:{from}\:{first}\:{principle} \\ $$$${y}=\frac{\mathrm{1}}{{x}^{\mathrm{2}} +\mathrm{5}} \\ $$
Answered by bobhans last updated on 29/Apr/22
![(dy/dx) = lim_(p→0) ((f(x+p)−f(x))/p) = lim_(p→0) (((1/((x+p)^2 +5)) −(1/(x^2 +5)))/p) = lim_(p→0) (((x^2 +5)−(x^2 +2px+p^2 +5))/(p(x^2 +5)[(x+p)^2 +5 ])) = lim_(p→0) (1/((x^2 +5)[(x+p)^2 +5 ])) .lim_(p→0) ((−2px−p^2 )/p) = (1/((x^2 +5)^2 )) .lim_(p→0) ((−2x−p)/1) = ((−2x)/((x^2 +5)^2 ))](https://www.tinkutara.com/question/Q169361.png)
$$\:\:\frac{{dy}}{{dx}}\:=\:\underset{{p}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{{f}\left({x}+{p}\right)−{f}\left({x}\right)}{{p}} \\ $$$$\:\:\:\:\:\:\:\:\:=\:\underset{{p}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{\frac{\mathrm{1}}{\left({x}+{p}\right)^{\mathrm{2}} +\mathrm{5}}\:−\frac{\mathrm{1}}{{x}^{\mathrm{2}} +\mathrm{5}}}{{p}} \\ $$$$\:\:\:\:\:\:\:\:=\:\underset{{p}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{\left({x}^{\mathrm{2}} +\mathrm{5}\right)−\left({x}^{\mathrm{2}} +\mathrm{2}{px}+{p}^{\mathrm{2}} +\mathrm{5}\right)}{{p}\left({x}^{\mathrm{2}} +\mathrm{5}\right)\left[\left({x}+{p}\right)^{\mathrm{2}} +\mathrm{5}\:\right]} \\ $$$$\:\:\:\:\:\:=\:\underset{{p}\rightarrow\mathrm{0}} {\mathrm{lim}}\frac{\mathrm{1}}{\left({x}^{\mathrm{2}} +\mathrm{5}\right)\left[\left({x}+{p}\right)^{\mathrm{2}} +\mathrm{5}\:\right]}\:.\underset{{p}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{−\mathrm{2}{px}−{p}^{\mathrm{2}} }{{p}} \\ $$$$\:\:\:\:=\:\frac{\mathrm{1}}{\left({x}^{\mathrm{2}} +\mathrm{5}\right)^{\mathrm{2}} }\:.\underset{{p}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{−\mathrm{2}{x}−{p}}{\mathrm{1}}\:=\:\frac{−\mathrm{2}{x}}{\left({x}^{\mathrm{2}} +\mathrm{5}\right)^{\mathrm{2}} }\: \\ $$