# How do you differentiate f(x)=(x^4-1)(e^x-2) using the product rule?

Feb 21, 2016

You must apply the chain rule alongside some derivative results
$\frac{d \left({x}^{4} - 1\right) \left({e}^{x} - 2\right)}{\mathrm{dx}} = 4 {x}^{3} \left({e}^{x} - 2\right) + \left({x}^{4} - 1\right) {e}^{x}$

#### Explanation:

The chain rule reads, for two given functions $f \left(x\right)$ and $g \left(x\right)$ :

$\frac{\mathrm{df} \left(x\right) g \left(x\right)}{\mathrm{dx}} = \frac{\mathrm{df} \left(x\right)}{\mathrm{dx}} g \left(x\right) + f \left(x\right) \frac{\mathrm{dg} \left(x\right)}{\mathrm{dx}}$

IMP . in case your functions are not a function of the independent variables directly, e.g. you make the derivative regarding the time, not x, you must pay attention to that, some previous maneuver must be done.

So:

$\frac{d \left({x}^{4} - 1\right) \left({e}^{x} - 2\right)}{\mathrm{dx}} = \frac{d \left({x}^{4} - 1\right)}{\mathrm{dx}} \left({e}^{x} - 2\right) + \left({x}^{4} - 1\right) \frac{d \left({e}^{x} - 2\right)}{\mathrm{dx}}$

Now you must solve the derivatives one by one:

$\frac{d \left({x}^{4} - 1\right)}{\mathrm{dx}}$

Use the linearity property of derivative, the derivative of the sum is the sum of the derivatives, it is a linear operator. Then remember that the derivative of polynomial is just the subtraction of the exponent, multiplied by the previous exponent.

Finally, remember that the derivative of a constant is zero.

$\frac{d \left({x}^{4} - 1\right)}{\mathrm{dx}} = 4 {x}^{3}$

For the exponential, just remember that the derivative of the exponential is itself.

Finally you get:

$\frac{d \left({x}^{4} - 1\right) \left({e}^{x} - 2\right)}{\mathrm{dx}} = 4 {x}^{3} \left({e}^{x} - 2\right) + \left({x}^{4} - 1\right) {e}^{x}$

If you want you can simply the expression, but for now it is not needed.