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Given y = [tex]\frac{2x-5}{x^{2} -2}[/tex], find the value of [tex]\frac{dy}{dx}[/tex] at x = 2.

Sagot :

Kailes

Given:

[tex]\bold{\dfrac{2x-5}{x^{2} -2}}[/tex]

[tex]\bold{\dfrac{dy}{dx}}[/tex]

[tex] \huge\mathbb{ \underline{SOLUTION :}}[/tex]

» [tex] \tt{For \: \: y,}[/tex]

[tex]\longrightarrow\sf{y=v \dfrac{2x - 5}{ {x}^{2} - 2} }[/tex]

[tex]\longrightarrow\sf{\dfrac{dy}{dx} = \cfrac{ ( {x}^{2} - 2)(2) - (2x - 5)(2x)}{( {x}^{2} - {2}^{2} )}}[/tex]

[tex]\longrightarrow\sf{\cfrac{2 {x}^{2} - 4 - {4x}^{2} + 10x}{ ({x}^{2} - {2}^{2} )}}[/tex]

[tex]\longrightarrow{={ \boxed{\sf \cfrac{ - 2 {x}^{2} - 4 + 10x}{ ({x}^{2} - {2}^{2} )}}}}[/tex]

» [tex] \tt{At \: \: x = 2,}[/tex]

[tex]\longrightarrow\sf{ \dfrac{ - 2(2) {}^{2} + 10(2) - 4 }{( {2}^{2} - 2 {)}^{2} } }[/tex]

[tex]\longrightarrow\sf{\dfrac{ - 8 + 20 - 4}{4} }[/tex]

[tex]\longrightarrow\sf{ \dfrac{8}{4} }[/tex]

[tex]\longrightarrow{\sf = \boxed{\sf {2}}}[/tex]

[tex]\huge \mathbb{ \underline{ANSWER:}}[/tex]

The value of the given differential function at [tex]\sf{x=2}[/tex] is [tex]\sf{2.}[/tex]