Question Number 59720 by Tawa1 last updated on 13/May/19
$$\mathrm{Find}\:\:\frac{\mathrm{dy}}{\mathrm{dx}}\:\:\mathrm{from}\:\mathrm{first}\:\mathrm{principle},\:\:\mathrm{if}\:\:\:\:\mathrm{y}\:=\:\mathrm{sin}^{\mathrm{2}} \left(\mathrm{x}\right) \\ $$
Commented by maxmathsup by imad last updated on 14/May/19
$$\frac{{dy}}{{dx}}\:=\mathrm{2}{sinx}\:{cosx}\:\:={sin}\left(\mathrm{2}{x}\right). \\ $$
Answered by tanmay last updated on 14/May/19
![y=sin^2 x y+△y=sin^2 (x+△x) △y=sin^2 (x+△x)−sin^2 x △y=[sin(x+△x)+sinx]×[sin(x+△x)−sinx] △y=2sin(((x+△x+x)/2))×cos(((△x)/2))×2cos(((x+△x+x)/2))sin(((△x)/2)) △y=2sin(2x+△x)cos(((△x)/2))sin(((△x)/2)) (dy/dx)=lim_(△x→0) ((△y)/(△x)) =lim_(△x→0) ((2sin(2x+△x)cos(((△x)/2))×sin(((△x)/2)))/(△x)) =lim_(△x→0) sin(2x+△x)cos(((△x)/2))×((sin(((△x)/2)))/((△x)/2)) =sin(2x) [lim_(△x→0) cos(((△x)/2))=1 and lim_(△x→0) ((sin(((△x)/2)))/((△x)/2))=1]](https://www.tinkutara.com/question/Q59737.png)
$${y}={sin}^{\mathrm{2}} {x} \\ $$$${y}+\bigtriangleup{y}={sin}^{\mathrm{2}} \left({x}+\bigtriangleup{x}\right) \\ $$$$\bigtriangleup{y}={sin}^{\mathrm{2}} \left({x}+\bigtriangleup{x}\right)−{sin}^{\mathrm{2}} {x} \\ $$$$\bigtriangleup{y}=\left[{sin}\left({x}+\bigtriangleup{x}\right)+{sinx}\right]×\left[{sin}\left({x}+\bigtriangleup{x}\right)−{sinx}\right] \\ $$$$\bigtriangleup{y}=\mathrm{2}{sin}\left(\frac{{x}+\bigtriangleup{x}+{x}}{\mathrm{2}}\right)×{cos}\left(\frac{\bigtriangleup{x}}{\mathrm{2}}\right)×\mathrm{2}{cos}\left(\frac{{x}+\bigtriangleup{x}+{x}}{\mathrm{2}}\right){sin}\left(\frac{\bigtriangleup{x}}{\mathrm{2}}\right) \\ $$$$\bigtriangleup{y}=\mathrm{2}{sin}\left(\mathrm{2}{x}+\bigtriangleup{x}\right){cos}\left(\frac{\bigtriangleup{x}}{\mathrm{2}}\right){sin}\left(\frac{\bigtriangleup{x}}{\mathrm{2}}\right) \\ $$$$\frac{{dy}}{{dx}}=\underset{\bigtriangleup{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\:\:\frac{\bigtriangleup{y}}{\bigtriangleup{x}} \\ $$$$=\underset{\bigtriangleup{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\:\:\frac{\mathrm{2}{sin}\left(\mathrm{2}{x}+\bigtriangleup{x}\right){cos}\left(\frac{\bigtriangleup{x}}{\mathrm{2}}\right)×{sin}\left(\frac{\bigtriangleup{x}}{\mathrm{2}}\right)}{\bigtriangleup{x}} \\ $$$$=\underset{\bigtriangleup{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\:\:{sin}\left(\mathrm{2}{x}+\bigtriangleup{x}\right){cos}\left(\frac{\bigtriangleup{x}}{\mathrm{2}}\right)×\frac{{sin}\left(\frac{\bigtriangleup{x}}{\mathrm{2}}\right)}{\frac{\bigtriangleup{x}}{\mathrm{2}}} \\ $$$$={sin}\left(\mathrm{2}{x}\right)\:\:\:\left[\underset{\bigtriangleup{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\:{cos}\left(\frac{\bigtriangleup{x}}{\mathrm{2}}\right)=\mathrm{1}\:\:\:{and}\:\underset{\bigtriangleup{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{{sin}\left(\frac{\bigtriangleup{x}}{\mathrm{2}}\right)}{\frac{\bigtriangleup{x}}{\mathrm{2}}}=\mathrm{1}\right] \\ $$
Commented by Tawa1 last updated on 14/May/19
$$\mathrm{God}\:\mathrm{bless}\:\mathrm{you}\:\mathrm{sir} \\ $$
Commented by tanmay last updated on 14/May/19
$${thank}\:{you}… \\ $$
Answered by Kunal12588 last updated on 14/May/19
![y=sin^2 x ⇒Δy=sin^2 (x+Δx)−sin^2 x ⇒Δy=((1−cos(2x+2Δx))/2)−((1−cos(2x))/2) [∵ cos 2θ =1−2sin^2 θ ⇒ sin^2 θ=((1−cos 2θ)/2) ] ⇒Δy=((cos (2x) − cos (2x+2Δx))/2) ⇒Δy=((2sin((4x+2Δx)/2)sin((2Δx)/2))/2) ⇒((Δy)/(Δx))=((sin(2x+Δx)sinΔx)/(Δx)) ⇒lim_(Δx→0) ((Δy)/(Δx)) = lim_(Δx→0) {sin(2x+Δx) ((sinΔx)/(Δx))} ⇒(dy/dx)=sin (2x)](https://www.tinkutara.com/question/Q59760.png)
$${y}={sin}^{\mathrm{2}} {x} \\ $$$$\Rightarrow\Delta{y}={sin}^{\mathrm{2}} \left({x}+\Delta{x}\right)−{sin}^{\mathrm{2}} {x} \\ $$$$\Rightarrow\Delta{y}=\frac{\mathrm{1}−{cos}\left(\mathrm{2}{x}+\mathrm{2}\Delta{x}\right)}{\mathrm{2}}−\frac{\mathrm{1}−{cos}\left(\mathrm{2}{x}\right)}{\mathrm{2}} \\ $$$$\left[\because\:{cos}\:\mathrm{2}\theta\:=\mathrm{1}−\mathrm{2}{sin}^{\mathrm{2}} \theta\:\Rightarrow\:{sin}^{\mathrm{2}} \theta=\frac{\mathrm{1}−{cos}\:\mathrm{2}\theta}{\mathrm{2}}\:\right] \\ $$$$\Rightarrow\Delta{y}=\frac{{cos}\:\left(\mathrm{2}{x}\right)\:−\:{cos}\:\left(\mathrm{2}{x}+\mathrm{2}\Delta{x}\right)}{\mathrm{2}} \\ $$$$\Rightarrow\Delta{y}=\frac{\mathrm{2}{sin}\frac{\mathrm{4}{x}+\mathrm{2}\Delta{x}}{\mathrm{2}}{sin}\frac{\mathrm{2}\Delta{x}}{\mathrm{2}}}{\mathrm{2}} \\ $$$$\Rightarrow\frac{\Delta{y}}{\Delta{x}}=\frac{{sin}\left(\mathrm{2}{x}+\Delta{x}\right){sin}\Delta{x}}{\Delta{x}} \\ $$$$\Rightarrow\underset{\Delta{x}\rightarrow\mathrm{0}} {{lim}}\:\frac{\Delta{y}}{\Delta{x}}\:=\:\underset{\Delta{x}\rightarrow\mathrm{0}} {{lim}}\:\left\{{sin}\left(\mathrm{2}{x}+\Delta{x}\right)\:\frac{{sin}\Delta{x}}{\Delta{x}}\right\} \\ $$$$\Rightarrow\frac{{dy}}{{dx}}={sin}\:\left(\mathrm{2}{x}\right) \\ $$
Commented by Tawa1 last updated on 19/May/19
$$\mathrm{God}\:\mathrm{bless}\:\mathrm{you}\:\mathrm{sir} \\ $$