[tex]\underline{\sf{Solution}} \: : [/tex]
[tex]\bf{We \: have \: to \: find \: derivative \: by \: first \: principal \: of \: : }[/tex]
[tex]\longrightarrow\sf{ \dfrac{1}{\sqrt{2x + 3}}} [/tex]
[tex] \sf { Let \: the \: function \: be \: f(x) = \dfrac{1}{\sqrt{2x + 3}}} [/tex]
[tex]\therefore\sf{f(x + h) = \dfrac{1}{\sqrt{2(x + h) + 3}}} [/tex]
[tex]\implies\sf{f(x + h) = \dfrac{1}{\sqrt{2x + 2h + 3}}} [/tex]
[tex]\sf We \: know, \: using \: 1st \: principal \: of \: derivation \: formula \: : [/tex]
[tex]\underline{\boxed{\bold{ \dfrac{d}{dx} \: \big( f(x) \big) = \lim_{h \to 0} \: \dfrac{ f(x + h) - f(x)}{h} }}}[/tex]
[tex]\sf Now \: substituting \: all : [/tex]
[tex]\longrightarrow\sf{ \dfrac{d}{dx} \: \bigg(\dfrac{1}{\sqrt{2x + 3}} \bigg) = \lim_{h \to 0} \: \dfrac{ \dfrac{1}{\sqrt{2x + 2h + 3}} - \dfrac{1}{\sqrt{2x + 3}}}{h}} [/tex]
[tex] = \sf{ \lim_{h \to 0} \dfrac{ \dfrac{ \sqrt{2x + 3} - \sqrt{2x + 2h + 3} }{ \sqrt{(2x + 3)(2x + 2h + 3)}}} {h} }[/tex]
[tex] = \sf{\lim_{h \to 0} \dfrac{1}{h \sqrt{(2x + 3)(2x + 2h + 3)}} \: ( \sqrt{2x + 3} - \sqrt{2x + 2h + 3})}[/tex]
[tex] = \sf{ \dfrac{1}{\sqrt{(2x + 3)(2x + 0 + 3)}} \: \lim_{h \to 0} \dfrac{ ( \sqrt{2x + 3} - \sqrt{2x + 2h + 3})( \sqrt{2x + 3} + \sqrt{2x + 2h + 3})}{ h(\sqrt{2x + 3} + \sqrt{2x + 2h + 3})}}[/tex]
[tex] = \sf{ \dfrac{1}{ \sqrt{ (2x + 3)(2x + 3)}} \: \lim_{h \to 0} \dfrac{ (\sqrt{2x + 3})^2 - (\sqrt{2x + 2h + 3})^2}{ h(\sqrt{2x + 3} + \sqrt{2x + 2h + 3}) }} [/tex]
[tex]= \sf{ \dfrac{1}{2x + 3} \: \lim_{h \to 0} \: \dfrac{2x + 3 - 2x - 2h - 3}{ h(\sqrt{2x + 3} + \sqrt{2x + 2h + 3}) }} [/tex]
[tex] = \sf{ \dfrac{1}{ 2x + 3 } \: \lim_{h \to 0} \: \dfrac{-2h}{h( \sqrt{2x + 3} + \sqrt{2x + 0 + 3})}} [/tex]
[tex] = \sf{\dfrac{1}{2x + 3} \times \dfrac{-2}{ \sqrt{2x + 3} + \sqrt{2x + 3} }}[/tex]
[tex] = \sf{\dfrac{1}{2x + 3} \times \dfrac{-2}{2\sqrt{2x + 3} }} [/tex]
[tex] = \sf{\dfrac{1}{2x + 3} \times \dfrac{-1}{\sqrt{2x + 3}}}[/tex]
[tex] = \sf{\dfrac{-1}{(2x + 3) \sqrt{2x + 3}}} [/tex]
[tex]= \underline{\boxed{\bold{ \dfrac{-1}{ (2x + 3)^{3/2} }}}}[/tex]
[tex]\sf{Therefore,}[/tex]
[tex]\sf{Derivation \: of \: given \: function = \boxed{ \dfrac{-1}{ (2x + 3)^{3/2}}} }[/tex]
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Answers & Comments
[tex]\underline{\sf{Solution}} \: : [/tex]
[tex]\bf{We \: have \: to \: find \: derivative \: by \: first \: principal \: of \: : }[/tex]
[tex]\longrightarrow\sf{ \dfrac{1}{\sqrt{2x + 3}}} [/tex]
[tex] \sf { Let \: the \: function \: be \: f(x) = \dfrac{1}{\sqrt{2x + 3}}} [/tex]
[tex]\therefore\sf{f(x + h) = \dfrac{1}{\sqrt{2(x + h) + 3}}} [/tex]
[tex]\implies\sf{f(x + h) = \dfrac{1}{\sqrt{2x + 2h + 3}}} [/tex]
[tex]\sf We \: know, \: using \: 1st \: principal \: of \: derivation \: formula \: : [/tex]
[tex]\underline{\boxed{\bold{ \dfrac{d}{dx} \: \big( f(x) \big) = \lim_{h \to 0} \: \dfrac{ f(x + h) - f(x)}{h} }}}[/tex]
[tex]\sf Now \: substituting \: all : [/tex]
[tex]\longrightarrow\sf{ \dfrac{d}{dx} \: \bigg(\dfrac{1}{\sqrt{2x + 3}} \bigg) = \lim_{h \to 0} \: \dfrac{ \dfrac{1}{\sqrt{2x + 2h + 3}} - \dfrac{1}{\sqrt{2x + 3}}}{h}} [/tex]
[tex] = \sf{ \lim_{h \to 0} \dfrac{ \dfrac{ \sqrt{2x + 3} - \sqrt{2x + 2h + 3} }{ \sqrt{(2x + 3)(2x + 2h + 3)}}} {h} }[/tex]
[tex] = \sf{\lim_{h \to 0} \dfrac{1}{h \sqrt{(2x + 3)(2x + 2h + 3)}} \: ( \sqrt{2x + 3} - \sqrt{2x + 2h + 3})}[/tex]
[tex] = \sf{ \dfrac{1}{\sqrt{(2x + 3)(2x + 0 + 3)}} \: \lim_{h \to 0} \dfrac{ ( \sqrt{2x + 3} - \sqrt{2x + 2h + 3})( \sqrt{2x + 3} + \sqrt{2x + 2h + 3})}{ h(\sqrt{2x + 3} + \sqrt{2x + 2h + 3})}}[/tex]
[tex] = \sf{ \dfrac{1}{ \sqrt{ (2x + 3)(2x + 3)}} \: \lim_{h \to 0} \dfrac{ (\sqrt{2x + 3})^2 - (\sqrt{2x + 2h + 3})^2}{ h(\sqrt{2x + 3} + \sqrt{2x + 2h + 3}) }} [/tex]
[tex]= \sf{ \dfrac{1}{2x + 3} \: \lim_{h \to 0} \: \dfrac{2x + 3 - 2x - 2h - 3}{ h(\sqrt{2x + 3} + \sqrt{2x + 2h + 3}) }} [/tex]
[tex] = \sf{ \dfrac{1}{ 2x + 3 } \: \lim_{h \to 0} \: \dfrac{-2h}{h( \sqrt{2x + 3} + \sqrt{2x + 0 + 3})}} [/tex]
[tex] = \sf{\dfrac{1}{2x + 3} \times \dfrac{-2}{ \sqrt{2x + 3} + \sqrt{2x + 3} }}[/tex]
[tex] = \sf{\dfrac{1}{2x + 3} \times \dfrac{-2}{2\sqrt{2x + 3} }} [/tex]
[tex] = \sf{\dfrac{1}{2x + 3} \times \dfrac{-1}{\sqrt{2x + 3}}}[/tex]
[tex] = \sf{\dfrac{-1}{(2x + 3) \sqrt{2x + 3}}} [/tex]
[tex]= \underline{\boxed{\bold{ \dfrac{-1}{ (2x + 3)^{3/2} }}}}[/tex]
[tex]\sf{Therefore,}[/tex]
[tex]\sf{Derivation \: of \: given \: function = \boxed{ \dfrac{-1}{ (2x + 3)^{3/2}}} }[/tex]