The derivative of the function $\frac{1}{x}$ is definitively calculated as $\frac{-1}{x^2}$. The limit definition of the derivative, expressed as $\frac{1/(x+h)-1/x}{h}$, simplifies to $\frac{-1}{x(x+h)}$. Plugging in $h=0$ yields the correct derivative, confirming that the limit approaches the expected result. The discussion highlights the importance of understanding the non-associative nature of division in calculus.
PREREQUISITESStudents of calculus, mathematics educators, and anyone seeking to deepen their understanding of derivatives and limit processes.
Fermat said:I want to show the derivative of $\frac{1}{x}$ is $\frac{-1}{x^2}$. Well, $\frac{1/(x+h)-1/x}{h}=\frac{-h^2}{x(x+h)}$. Why can't I just plug in $h=0$ to get that the limit is 0?
I like Serena said:$$\frac{1/(x+h)-1/x}{h}=\frac{-1}{x(x+h)}$$
The value $h=0$ can indeed be plugged in and gives the desired result.