Question Number 60475 by Tawa1 last updated on 21/May/19
Answered by tanmay last updated on 21/May/19
)/dx))/({f(x)}^2 )) p′(2)=((f(2)g′(2)−[g(2)−1]f′(2))/(f^2 (2))) =(((−1)(−sin2)−[cos2−1]5)/((−1)^2 )) =sin2−5cos2+5 =7.99](https://www.tinkutara.com/question/Q60482.png)
$$\frac{{dp}}{{dx}}=\frac{{f}\left({x}\right)\frac{{dg}\left({x}\right)}{{dx}}−\left[{g}\left({x}\right)−\mathrm{1}\right]\frac{{df}\left({x}\right)}{{dx}}}{\left\{{f}\left({x}\right)\right\}^{\mathrm{2}} } \\ $$$${p}'\left(\mathrm{2}\right)=\frac{{f}\left(\mathrm{2}\right){g}'\left(\mathrm{2}\right)−\left[{g}\left(\mathrm{2}\right)−\mathrm{1}\right]{f}'\left(\mathrm{2}\right)}{{f}^{\mathrm{2}} \left(\mathrm{2}\right)} \\ $$$$=\frac{\left(−\mathrm{1}\right)\left(−{sin}\mathrm{2}\right)−\left[{cos}\mathrm{2}−\mathrm{1}\right]\mathrm{5}}{\left(−\mathrm{1}\right)^{\mathrm{2}} } \\ $$$$={sin}\mathrm{2}−\mathrm{5}{cos}\mathrm{2}+\mathrm{5} \\ $$$$=\mathrm{7}.\mathrm{99} \\ $$
Commented by Tawa1 last updated on 21/May/19
$$\mathrm{God}\:\mathrm{bless}\:\mathrm{you}\:\mathrm{sir} \\ $$