Question

Given y = 2}" alt="\frac{2x-5}{x^{2} -2}" align="absmiddle" class="latex-formula">, find the value of $\frac{dy}{dx}$ at x = 2.

Answer 1

Given:

▪ $\bold{\dfrac{2x-5}{x^{2} -2}}$

▪ $\bold{\dfrac{dy}{dx}}$

$\huge\mathbb{ \underline{SOLUTION :}}$

» $\tt{For \: \: y,}$

$\longrightarrow\sf{y=v \dfrac{2x - 5}{ {x}^{2} - 2} }$

$\longrightarrow\sf{\dfrac{dy}{dx} = \cfrac{ ( {x}^{2} - 2)(2) - (2x - 5)(2x)}{( {x}^{2} - {2}^{2} )}}$

$\longrightarrow\sf{\cfrac{2 {x}^{2} - 4 - {4x}^{2} + 10x}{ ({x}^{2} - {2}^{2} )}}$

$\longrightarrow{={ \boxed{\sf \cfrac{ - 2 {x}^{2} - 4 + 10x}{ ({x}^{2} - {2}^{2} )}}}}$

» $\tt{At \: \: x = 2,}$

$\longrightarrow\sf{ \dfrac{ - 2(2) {}^{2} + 10(2) - 4 }{( {2}^{2} - 2 {)}^{2} } }$

$\longrightarrow\sf{\dfrac{ - 8 + 20 - 4}{4} }$

$\longrightarrow\sf{ \dfrac{8}{4} }$

$\longrightarrow{\sf = \boxed{\sf {2}}}$ ✓

$\huge \mathbb{ \underline{ANSWER:}}$

◆ The value of the given differential function at $\sf{x=2}$ is $\sf{2.}$