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.

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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 $\Delta x_0 \to 0$.

Answer

$\Delta y = \frac{61}{64} = 0.953125$A, $dy = \frac{3}{4} = 0.75$A.