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:\(\displaystyle \frac{1/(x+h)-1/x}{h}=\frac{-1}{x(x+h)}\)
The value $h=0$ can indeed be plugged in and gives the desired result.
A derivative is a mathematical concept that measures the instantaneous rate of change of a function at any given point. It is denoted by the symbol 'f'(x) or dy/dx.
The derivative of a fraction is calculated using the power rule, which states that the derivative of a function raised to a constant power is equal to the constant times the original function raised to the power minus one. In the case of $\frac{1}{x}$, the derivative is $\frac{-1}{x^2}$.
The derivative of a function represents the slope of the tangent line at any given point on the graph of that function. In the case of $\frac{1}{x}$, the slope of the tangent line is constantly changing, because the function approaches zero at different rates for different values of x. Therefore, the derivative is negative and inversely proportional to x, resulting in $\frac{-1}{x^2}$.
The negative sign in the derivative of $\frac{1}{x}$ indicates that the function is decreasing at an increasing rate. This is because as x increases, $\frac{1}{x}$ decreases at a faster rate, resulting in a negative slope for the tangent line.
The limit definition of a derivative is used to find the derivative of a function at a specific point. To find the derivative of $\frac{1}{x}$ using this method, we start by writing out the limit definition: $f'(x) = \lim_{h\to 0} \frac{f(x+h)-f(x)}{h}$. Then, we substitute $\frac{1}{x}$ for f(x) and simplify the equation to get $f'(x) = \frac{-x}{x^2h}$. Finally, we take the limit as h approaches 0 to get the derivative of $\frac{1}{x}$, which is $\frac{-1}{x^2}$.