Find (dy/dx) from first principle, if y = sin^2 (x)


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Question Number 59720 by Tawa1 last updated on 13/May/19

$Find \frac{dy}{dx} from first principle, if y = \sin^{2} (x)$
Commented by maxmathsup by imad last updated on 14/May/19

$\frac{d y}{d x} = 2 s i n x c o s x = s i n (2 x) .$
	Answered by tanmay last updated on 14/May/19

	$y = {s i n}^{2} x$ $y + △ y = {s i n}^{2} (x + △ x)$ $△ y = {s i n}^{2} (x + △ x) - {s i n}^{2} x$ $△ y = [s i n (x + △ x) + s i n x] \times [s i n (x + △ x) - s i n x]$ $△ y = 2 s i n (\frac{x + △ x + x}{2}) \times c o s (\frac{△ x}{2}) \times 2 c o s (\frac{x + △ x + x}{2}) s i n (\frac{△ x}{2})$ $△ y = 2 s i n (2 x + △ x) c o s (\frac{△ x}{2}) s i n (\frac{△ x}{2})$ $\frac{d y}{d x} = \lim_{△ x \to 0} \frac{△ y}{△ x}$ $= \lim_{△ x \to 0} \frac{2 s i n (2 x + △ x) c o s (\frac{△ x}{2}) \times s i n (\frac{△ x}{2})}{△ x}$ $= \lim_{△ x \to 0} s i n (2 x + △ x) c o s (\frac{△ x}{2}) \times \frac{s i n (\frac{△ x}{2})}{\frac{△ x}{2}}$ $= s i n (2 x) [\lim_{△ x \to 0} c o s (\frac{△ x}{2}) = 1 a n d \lim_{△ x \to 0} \frac{s i n (\frac{△ x}{2})}{\frac{△ x}{2}} = 1]$
	Commented by Tawa1 last updated on 14/May/19

	$God bless you sir$
	Commented by tanmay last updated on 14/May/19

	$t h a n k y o u . . .$
	Answered by Kunal12588 last updated on 14/May/19
	$y=sin^2 x ⇒Δy=sin^2 (x+Δx)−sin^2 x ⇒Δy=((1−cos(2x+2Δx))/2)−((1−cos(2x))/2) [∵ cos 2θ =1−2sin^2 θ ⇒ sin^2 θ=((1−cos 2θ)/2) ] ⇒Δy=((cos (2x) − cos (2x+2Δx))/2) ⇒Δy=((2sin((4x+2Δx)/2)sin((2Δx)/2))/2) ⇒((Δy)/(Δx))=((sin(2x+Δx)sinΔx)/(Δx)) ⇒lim_(Δx→0) ((Δy)/(Δx)) = lim_(Δx→0) {sin(2x+Δx) ((sinΔx)/(Δx))} ⇒(dy/dx)=sin (2x)$
	$y = {s i n}^{2} x$ $\Rightarrow Δ y = {s i n}^{2} (x + Δ x) - {s i n}^{2} x$ $\Rightarrow Δ y = \frac{1 - c o s (2 x + 2 Δ x)}{2} - \frac{1 - c o s (2 x)}{2}$ $[∵ c o s 2 θ = 1 - 2 {s i n}^{2} θ \Rightarrow {s i n}^{2} θ = \frac{1 - c o s 2 θ}{2}]$ $\Rightarrow Δ y = \frac{c o s (2 x) - c o s (2 x + 2 Δ x)}{2}$ $\Rightarrow Δ y = \frac{2 s i n \frac{4 x + 2 Δ x}{2} s i n \frac{2 Δ x}{2}}{2}$ $\Rightarrow \frac{Δ y}{Δ x} = \frac{s i n (2 x + Δ x) s i n Δ x}{Δ x}$ $\Rightarrow \underset{Δ x \to 0}{l i m} \frac{Δ y}{Δ x} = \underset{Δ x \to 0}{l i m} {s i n (2 x + Δ x) \frac{s i n Δ x}{Δ x}}$ $\Rightarrow \frac{d y}{d x} = s i n (2 x)$
	Commented by Tawa1 last updated on 19/May/19

	$God bless you sir$
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