Question Number 54588 by hassentimol last updated on 07/Feb/19
![What is : (d/dx) [ u(x) × v(x) × w(x) ] = ... and more generally, what is : (d/dx) [ Π_(i=1) ^n u_i (x)] = ... Thank you](https://www.tinkutara.com/question/Q54588.png)
$$\mathrm{What}\:\mathrm{is}\:: \\ $$$$\:\:\:\frac{\mathrm{d}}{\mathrm{dx}}\:\left[\:{u}\left({x}\right)\:×\:{v}\left({x}\right)\:×\:{w}\left({x}\right)\:\right]\:=\:… \\ $$$$ \\ $$$$\mathrm{and}\:\mathrm{more}\:\mathrm{generally},\:\mathrm{what}\:\mathrm{is}\:: \\ $$$$\:\:\:\frac{\mathrm{d}}{\mathrm{dx}}\:\left[\:\underset{{i}=\mathrm{1}} {\overset{{n}} {\prod}}\:{u}_{{i}} \left({x}\right)\right]\:=\:… \\ $$$$ \\ $$$$\mathrm{Thank}\:\mathrm{you} \\ $$
Answered by tanmay.chaudhury50@gmail.com last updated on 07/Feb/19
![(d/dx)[Π_(i=1) ^n u_i (x)] let y=Π_(i=1) ^n u_i (x)=u_1 (x)u_2 (x)u_3 (x)...u_n (x) lny=lnu_1 (x)+lnu_2 (x)+lnu_3 (x)+...+lnu_n (x) (1/y)(dy/dx)=(1/(u_1 (x)))((du_1 (x))/dx)+(1/(u_2 (x)))((du_2 (x))/dx)+...+(1/(u_n (x)))((du_n (x))/dx) so (d/dx)[Π_(i=1) ^n u_i (x)]=[Π_(i=1) ^n u_i (x)]×[Σ_(i=1) ^n (1/(u_i (x)))((du_i (x))/dx)]](https://www.tinkutara.com/question/Q54589.png)
$$\frac{{d}}{{dx}}\left[\underset{{i}=\mathrm{1}} {\overset{{n}} {\prod}}{u}_{{i}} \left({x}\right)\right] \\ $$$${let}\:{y}=\underset{{i}=\mathrm{1}} {\overset{{n}} {\prod}}{u}_{{i}} \left({x}\right)={u}_{\mathrm{1}} \left({x}\right){u}_{\mathrm{2}} \left({x}\right){u}_{\mathrm{3}} \left({x}\right)…{u}_{{n}} \left({x}\right) \\ $$$${lny}={lnu}_{\mathrm{1}} \left({x}\right)+{lnu}_{\mathrm{2}} \left({x}\right)+{lnu}_{\mathrm{3}} \left({x}\right)+…+{lnu}_{{n}} \left({x}\right) \\ $$$$\frac{\mathrm{1}}{{y}}\frac{{dy}}{{dx}}=\frac{\mathrm{1}}{{u}_{\mathrm{1}} \left({x}\right)}\frac{{du}_{\mathrm{1}} \left({x}\right)}{{dx}}+\frac{\mathrm{1}}{{u}_{\mathrm{2}} \left({x}\right)}\frac{{du}_{\mathrm{2}} \left({x}\right)}{{dx}}+…+\frac{\mathrm{1}}{{u}_{{n}} \left({x}\right)}\frac{{du}_{{n}} \left({x}\right)}{{dx}} \\ $$$${so} \\ $$$$\frac{{d}}{{dx}}\left[\underset{{i}=\mathrm{1}} {\overset{{n}} {\prod}}{u}_{{i}} \left({x}\right)\right]=\left[\underset{{i}=\mathrm{1}} {\overset{{n}} {\prod}}{u}_{{i}} \left({x}\right)\right]×\left[\underset{{i}=\mathrm{1}} {\overset{{n}} {\sum}}\frac{\mathrm{1}}{{u}_{{i}} \left({x}\right)}\frac{{du}_{{i}} \left({x}\right)}{{dx}}\right] \\ $$
Commented by hassentimol last updated on 07/Feb/19
$$\mathrm{Thank}\:\mathrm{you}\:\mathrm{very}\:\mathrm{much} \\ $$
Commented by tanmay.chaudhury50@gmail.com last updated on 08/Feb/19
$${most}\:{welcome}… \\ $$