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

Question

Evelyn Agard · Accepted Answer

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. So: #(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.

Charles Arocha · Answer

undefined To differentiate $ f(x) = (x^4 - 1)(e^x - 2) $ using the product rule:

1. Apply the product rule, which states that the derivative of the product of two functions $ u(x) $ and $ v(x) $ is given by $ u'(x)v(x) + u(x)v'(x) $.
2. Let $ u(x) = x^4 - 1 $ and $ v(x) = e^x - 2 $.
3. Calculate the derivatives $ u'(x) $ and $ v'(x) $.
   - $ u'(x) = 4x^3 $ (using the power rule)
   - $ v'(x) = e^x $ (derivative of $ e^x $ is itself)
4. Apply the product rule formula:
   $ f'(x) = u'(x)v(x) + u(x)v'(x) $
   $ f'(x) = (4x^3)(e^x - 2) + (x^4 - 1)(e^x) $
5. Simplify the expression if necessary.