Instantaneous Rate of Change Calculator

Find the instantaneous rate of change of $$$x^{3} + 5 x^{2} + 7 x + 4$$$ at $$$x = 6$$$.

The instantaneous rate of change of the function $$$f{\left(x \right)}$$$ at the point $$$x = x_{0}$$$ is the derivative of the function $$$f{\left(x \right)}$$$ evaluated at the point $$$x = x_{0}$$$.

So, find the derivative of the function.

The derivative of a sum/difference is the sum/difference of derivatives:

$$\color{red}{\left(\frac{d}{dx} \left(x^{3} + 5 x^{2} + 7 x + 4\right)\right)} = \color{red}{\left(\frac{d}{dx} \left(x^{3}\right) + \frac{d}{dx} \left(5 x^{2}\right) + \frac{d}{dx} \left(7 x\right) + \frac{d}{dx} \left(4\right)\right)}$$

Apply the constant multiple rule $$$\frac{d}{dx} \left(c f{\left(x \right)}\right) = c \frac{d}{dx} \left(f{\left(x \right)}\right)$$$ with $$$c = 7$$$ and $$$f{\left(x \right)} = x$$$:

$$\color{red}{\left(\frac{d}{dx} \left(7 x\right)\right)} + \frac{d}{dx} \left(4\right) + \frac{d}{dx} \left(5 x^{2}\right) + \frac{d}{dx} \left(x^{3}\right) = \color{red}{\left(7 \frac{d}{dx} \left(x\right)\right)} + \frac{d}{dx} \left(4\right) + \frac{d}{dx} \left(5 x^{2}\right) + \frac{d}{dx} \left(x^{3}\right)$$

Apply the power rule $$$\frac{d}{dx} \left(x^{n}\right) = n x^{n - 1}$$$ with $$$n = 1$$$, in other words, $$$\frac{d}{dx} \left(x\right) = 1$$$:

$$7 \color{red}{\left(\frac{d}{dx} \left(x\right)\right)} + \frac{d}{dx} \left(4\right) + \frac{d}{dx} \left(5 x^{2}\right) + \frac{d}{dx} \left(x^{3}\right) = 7 \color{red}{\left(1\right)} + \frac{d}{dx} \left(4\right) + \frac{d}{dx} \left(5 x^{2}\right) + \frac{d}{dx} \left(x^{3}\right)$$

Apply the constant multiple rule $$$\frac{d}{dx} \left(c f{\left(x \right)}\right) = c \frac{d}{dx} \left(f{\left(x \right)}\right)$$$ with $$$c = 5$$$ and $$$f{\left(x \right)} = x^{2}$$$:

$$\color{red}{\left(\frac{d}{dx} \left(5 x^{2}\right)\right)} + \frac{d}{dx} \left(4\right) + \frac{d}{dx} \left(x^{3}\right) + 7 = \color{red}{\left(5 \frac{d}{dx} \left(x^{2}\right)\right)} + \frac{d}{dx} \left(4\right) + \frac{d}{dx} \left(x^{3}\right) + 7$$

Apply the power rule $$$\frac{d}{dx} \left(x^{n}\right) = n x^{n - 1}$$$ with $$$n = 3$$$:

$$\color{red}{\left(\frac{d}{dx} \left(x^{3}\right)\right)} + \frac{d}{dx} \left(4\right) + 5 \frac{d}{dx} \left(x^{2}\right) + 7 = \color{red}{\left(3 x^{2}\right)} + \frac{d}{dx} \left(4\right) + 5 \frac{d}{dx} \left(x^{2}\right) + 7$$

The derivative of a constant is $$$0$$$:

$$3 x^{2} + \color{red}{\left(\frac{d}{dx} \left(4\right)\right)} + 5 \frac{d}{dx} \left(x^{2}\right) + 7 = 3 x^{2} + \color{red}{\left(0\right)} + 5 \frac{d}{dx} \left(x^{2}\right) + 7$$

Apply the power rule $$$\frac{d}{dx} \left(x^{n}\right) = n x^{n - 1}$$$ with $$$n = 2$$$:

$$3 x^{2} + 5 \color{red}{\left(\frac{d}{dx} \left(x^{2}\right)\right)} + 7 = 3 x^{2} + 5 \color{red}{\left(2 x\right)} + 7$$

Thus, $$$\frac{d}{dx} \left(x^{3} + 5 x^{2} + 7 x + 4\right) = 3 x^{2} + 10 x + 7$$$.

Finally, evaluate the derivative at $$$x = 6$$$.

$$$\left(\frac{d}{dx} \left(x^{3} + 5 x^{2} + 7 x + 4\right)\right)|_{\left(x = 6\right)} = 175$$$

Therefore, the instantaneous rate of change of $$$x^{3} + 5 x^{2} + 7 x + 4$$$ at $$$x = 6$$$ is $$$175$$$.

The instantaneous rate of $$$x^{3} + 5 x^{2} + 7 x + 4$$$A at $$$x = 6$$$A is $$$175$$$A.