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

1 Answer
Feb 21, 2016

You must apply the chain rule alongside some derivative results
#(d(x^4 -1)(e^x -2))/dx=4x^3(e^x - 2) + (x^4 -1)e^x#


The chain rule reads, for two given functions #f(x)# and #g(x)# :

#(df(x)g(x))/dx= (df(x))/dxg(x) + f(x)(dg(x))/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.


#(d(x^4 -1)(e^x -2))/dx=(d(x^4 -1))/dx(e^x -2) + (x^4 -1)(d(e^x -2))/dx#

Now you must solve the derivatives one by one:

#(d(x^4 -1))/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.

#(d(x^4 -1))/dx=4x^3#

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

Finally you get:

#(d(x^4 -1)(e^x -2))/dx=4x^3(e^x - 2) + (x^4 -1)e^x#

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