# Differential of a Function Calculator

For the given function $$$y=f(x)$$$, point $$$x_0$$$ and argument change $$$\Delta x_0$$$, the calculator will find the differential $$$dy$$$ and the function change $$$\Delta y$$$, with steps shown.

## Your Input

**Find the differential $$$dy$$$ and the function change $$$\Delta y$$$ of $$$f{\left(x \right)} = x^{3}$$$ when $$$x_{0} = 1$$$ and $$$\Delta x_{0} = \frac{1}{4}$$$.**

## Solution

Find the second point: $$$x_{0} + \Delta x_{0} = 1 + \frac{1}{4} = \frac{5}{4}$$$.

Evaluate the function at the two points: $$$f{\left(x_{0} + \Delta x_{0} \right)} = f{\left(\frac{5}{4} \right)} = \frac{125}{64}$$$, $$$f{\left(x_{0} \right)} = f{\left(1 \right)} = 1$$$.

According to the definition: $$$\Delta y = f{\left(x_{0} + \Delta x_{0} \right)} - f{\left(x_{0} \right)} = \frac{125}{64} - 1 = \frac{61}{64}$$$.

Find the derivative: $$$f^{\prime }\left(x\right) = 3 x^{2}$$$ (for steps, see derivative calculator).

Evaluate the derivative at $$$x_{0} = 1$$$: $$$f^{\prime }\left(1\right) = 3$$$.

The differential is defined as $$$dy = f^{\prime }\left(x_{0}\right) \Delta x_{0} = \left(3\right)\cdot \left(\frac{1}{4}\right) = \frac{3}{4}$$$.

Note that the value of $$$dy$$$ becomes closer to $$$\Delta y$$$ as $$$\Delta x_0 \to 0$$$.