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Sagot :
SOLUTION
We want to differentiate
[tex]f\mleft(x\mright)=\frac{x^{2}-2x+5}{x^{2}}[/tex]Now fro quotient rule
[tex]f^{\prime}(x)=\frac{v\frac{du}{dx}-u\frac{dv}{dx}}{v^2}[/tex]From the question,
Let
[tex]\begin{gathered} u=x^2-2x+5 \\ \frac{du}{dx}=2x-2 \end{gathered}[/tex]And let
[tex]\begin{gathered} v=x^2 \\ \frac{dv}{dx}=2x \end{gathered}[/tex]Hence
[tex]\begin{gathered} f^{\prime}(x)=\frac{v\frac{du}{dx}-u\frac{dv}{dx}}{v^2} \\ f^{\prime}(x)=\frac{x^2(2x-2)-(x^2-2x+5)2x}{(x^2)^2} \\ f^{\prime}(x)=\frac{2x^3-2x^2-(2x^3-4x^2+10x)}{(x^2)^2} \\ f^{\prime}(x)=\frac{2x^3-2x^2-2x^3+4x^2-10x)}{x^4^{}} \\ f^{\prime}(x)=\frac{2x^2-10x}{x^4^{}} \\ f^{\prime}(x)=\frac{2x-10}{x^3^{}} \end{gathered}[/tex]Therefore, the answer is
[tex]f^{\prime}(x)=\frac{2x-10}{x^3^{}}[/tex]
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