Question

Differentiate f(x)=(x^2-2x+5)/(x^2)

Answer 1

We want to differentiate

`f\mleft(x\mright)=(x^(2)-2x+5)/(x^(2))`

Now fro quotient rule

`f^(\prime)(x)=(v(du)/(dx)-u(dv)/(dx))/(v^2)`

From the question,

Let

`\begin{gathered} u=x^2-2x+5 \\ (du)/(dx)=2x-2 \end{gathered}`

And let

`\begin{gathered} v=x^2 \\ (dv)/(dx)=2x \end{gathered}`

Hence

`\begin{gathered} f^(\prime)(x)=(v(du)/(dx)-u(dv)/(dx))/(v^2) \\ f^(\prime)(x)=(x^2(2x-2)-(x^2-2x+5)2x)/((x^2)^2) \\ f^(\prime)(x)=(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}`

Therefore, the answer is

`f^(\prime)(x)=\frac{2x-10}{x^3^{}}`