If y=[tex] ln( \sqrt{1 + x} ) – \sqrt{1 – x} ) \div ( \sqrt{1 + x } + \sqrt{1 – x} )[/tex]find dy/dx About the author Savannah
[tex]\begin{gathered}\Large{\bold{{\underline{Formula \: Used – }}}} \end{gathered}[/tex] [tex] \boxed{ \bf \: \dfrac{d}{dx} \: log(x) = \dfrac{1}{x} }[/tex] [tex] \boxed{ \bf \: \dfrac{d}{dx} \: x = 1}[/tex] [tex] \boxed{ \bf \: \dfrac{d}{dx} \:k = 0 }[/tex] [tex] \boxed{ \bf \: \dfrac{d}{dx} \: \sqrt{x} = \dfrac{1}{2 \sqrt{x} } }[/tex] [tex] \boxed{ \bf \: log(x) – log(y) = log\bigg( \dfrac{x}{y} \bigg) }[/tex] [tex]\large\underline{\sf{Solution-}}[/tex] [tex] \sf \:y = log\bigg( \dfrac{ \sqrt{1 + x} – \sqrt{1 – x} }{ \sqrt{1 + x} + \sqrt{1 – x} } \bigg) [/tex] [tex] \sf \:y = log\bigg( \dfrac{ \sqrt{1 + x} – \sqrt{1 – x} }{ \sqrt{1 + x} + \sqrt{1 – x}} \times \dfrac{ \sqrt{1 + x} – \sqrt{1 – x} }{ \sqrt{1 + x} – \sqrt{1 – x} } \bigg) [/tex] [tex] \sf \: y = log\bigg( \dfrac{ {\bigg( \sqrt{1 + x} – \sqrt{1 – x} \bigg) }^{2} }{ {\bigg( \sqrt{1 + x} \bigg) }^{2} – {\bigg( \sqrt{1 – x} \bigg) }^{2} } \bigg) [/tex] [tex] \sf \: y = log\bigg( \dfrac{1 + x + 1 – x – 2 \sqrt{1 – {x}^{2} } }{(1 + x) – (1 – x)} \bigg) [/tex] [tex] \sf \: y = log\bigg( \dfrac{2 – 2 \sqrt{1 – {x}^{2} } }{2x} \bigg) [/tex] [tex] \sf \: y = log\bigg( \dfrac{1 – \sqrt{1 – {x}^{2} } }{x} \bigg) [/tex] [tex] \sf \: y = log(1 – \sqrt{1 – {x}^{2} } ) – log(x) [/tex] On differentiating both sides, w. r. t. x [tex] \sf \: \dfrac{d}{dx} y = \dfrac{d}{dx} \bigg( log(1 – \sqrt{1 – {x}^{2} } ) – log(x) \bigg) [/tex] [tex] \sf \: \dfrac{dy}{dx} = \dfrac{1}{1 – \sqrt{1 – {x}^{2} } } \dfrac{d}{dx}(1 – \sqrt{1 – {x}^{2} }) – \dfrac{1}{x} [/tex] [tex] \sf \: \dfrac{dy}{dx} = \dfrac{1}{1 – \sqrt{1 – {x}^{2} } } \times \dfrac{( – 1)}{2 \sqrt{1 – {x}^{2} } } \dfrac{d}{dx}(1 – {x}^{2}) – \dfrac{1}{x} [/tex] [tex] \sf \: \dfrac{dy}{dx} = \dfrac{x}{(1 – \sqrt{1 – {x}^{2} }) \sqrt{1 – {x}^{2} } } – \dfrac{1}{x} [/tex] Additional Information :- [tex] \boxed{ \bf \: \dfrac{d}{dx} \: sinx = cosx}[/tex] [tex] \boxed{ \bf \: \dfrac{d}{dx} \:cosx = – \: sinx }[/tex] [tex] \boxed{ \bf \: \dfrac{d}{dx} \:tanx = {sec}^{2}x }[/tex] [tex] \boxed{ \bf \: \dfrac{d}{dx} \:cotx = – {cosec}^{2}x }[/tex] [tex] \boxed{ \bf \: \dfrac{d}{dx} \: {e}^{x} = {e}^{x} }[/tex] [tex] \boxed{ \bf \: \dfrac{d}{dx} \: u.v = u \: \dfrac{d}{dx}v \: + \: v \: \dfrac{d}{dx}u}[/tex] Reply
[tex]\begin{gathered}\Large{\bold{{\underline{Formula \: Used – }}}} \end{gathered}[/tex]
[tex] \boxed{ \bf \: \dfrac{d}{dx} \: log(x) = \dfrac{1}{x} }[/tex]
[tex] \boxed{ \bf \: \dfrac{d}{dx} \: x = 1}[/tex]
[tex] \boxed{ \bf \: \dfrac{d}{dx} \:k = 0 }[/tex]
[tex] \boxed{ \bf \: \dfrac{d}{dx} \: \sqrt{x} = \dfrac{1}{2 \sqrt{x} } }[/tex]
[tex] \boxed{ \bf \: log(x) – log(y) = log\bigg( \dfrac{x}{y} \bigg) }[/tex]
[tex]\large\underline{\sf{Solution-}}[/tex]
[tex] \sf \:y = log\bigg( \dfrac{ \sqrt{1 + x} – \sqrt{1 – x} }{ \sqrt{1 + x} + \sqrt{1 – x} } \bigg) [/tex]
[tex] \sf \:y = log\bigg( \dfrac{ \sqrt{1 + x} – \sqrt{1 – x} }{ \sqrt{1 + x} + \sqrt{1 – x}} \times \dfrac{ \sqrt{1 + x} – \sqrt{1 – x} }{ \sqrt{1 + x} – \sqrt{1 – x} } \bigg) [/tex]
[tex] \sf \: y = log\bigg( \dfrac{ {\bigg( \sqrt{1 + x} – \sqrt{1 – x} \bigg) }^{2} }{ {\bigg( \sqrt{1 + x} \bigg) }^{2} – {\bigg( \sqrt{1 – x} \bigg) }^{2} } \bigg) [/tex]
[tex] \sf \: y = log\bigg( \dfrac{1 + x + 1 – x – 2 \sqrt{1 – {x}^{2} } }{(1 + x) – (1 – x)} \bigg) [/tex]
[tex] \sf \: y = log\bigg( \dfrac{2 – 2 \sqrt{1 – {x}^{2} } }{2x} \bigg) [/tex]
[tex] \sf \: y = log\bigg( \dfrac{1 – \sqrt{1 – {x}^{2} } }{x} \bigg) [/tex]
[tex] \sf \: y = log(1 – \sqrt{1 – {x}^{2} } ) – log(x) [/tex]
On differentiating both sides, w. r. t. x
[tex] \sf \: \dfrac{d}{dx} y = \dfrac{d}{dx} \bigg( log(1 – \sqrt{1 – {x}^{2} } ) – log(x) \bigg) [/tex]
[tex] \sf \: \dfrac{dy}{dx} = \dfrac{1}{1 – \sqrt{1 – {x}^{2} } } \dfrac{d}{dx}(1 – \sqrt{1 – {x}^{2} }) – \dfrac{1}{x} [/tex]
[tex] \sf \: \dfrac{dy}{dx} = \dfrac{1}{1 – \sqrt{1 – {x}^{2} } } \times \dfrac{( – 1)}{2 \sqrt{1 – {x}^{2} } } \dfrac{d}{dx}(1 – {x}^{2}) – \dfrac{1}{x} [/tex]
[tex] \sf \: \dfrac{dy}{dx} = \dfrac{x}{(1 – \sqrt{1 – {x}^{2} }) \sqrt{1 – {x}^{2} } } – \dfrac{1}{x} [/tex]
Additional Information :-
[tex] \boxed{ \bf \: \dfrac{d}{dx} \: sinx = cosx}[/tex]
[tex] \boxed{ \bf \: \dfrac{d}{dx} \:cosx = – \: sinx }[/tex]
[tex] \boxed{ \bf \: \dfrac{d}{dx} \:tanx = {sec}^{2}x }[/tex]
[tex] \boxed{ \bf \: \dfrac{d}{dx} \:cotx = – {cosec}^{2}x }[/tex]
[tex] \boxed{ \bf \: \dfrac{d}{dx} \: {e}^{x} = {e}^{x} }[/tex]
[tex] \boxed{ \bf \: \dfrac{d}{dx} \: u.v = u \: \dfrac{d}{dx}v \: + \: v \: \dfrac{d}{dx}u}[/tex]