Find-dy-dx-from-first-principle-if-y-sin-2-x-

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 \dots

Answered by Kunal12588 last updated on 14/May/19

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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