How do you differentiate #f(x)=-xe^x*(4-x)/6# using the product rule?

Question

Christopher Agresta · Accepted Answer

#f'(x)=(xe^x)/6-(xe^x(4-x))/6-(e^x(4-x))/6# If a function #f(x)=uvw# where #u#, #v# and #w# are all functions of #x#, then #f'(x)=uvw'+uv'w+u'vw# #u=-x# #u'=-1# #v=e^x# #v'=e^x# #w=(4-x)/6=4/6-x/6# #w'=-1/6# #f'(x)=(xe^x)/6-(xe^x(4-x))/6-(e^x(4-x))/6#

Paisley Johnson · Answer

undefined To differentiate the function $ f(x) = -\frac{x e^x (4-x)}{6} $ using the product rule, you can follow these steps:

1. Identify the two functions being multiplied: $ u(x) = -\frac{x(4-x)}{6} $ and $ v(x) = e^x $.
2. Apply the product rule: $ f'(x) = u'(x)v(x) + u(x)v'(x) $.
3. Differentiate $ u(x) $ and $ v(x) $ separately.
4. Substitute the derivatives and the original functions into the product rule formula.

Here's how to do it step by step:

1. $ u(x) = -\frac{x(4-x)}{6} $ and $ v(x) = e^x $.
2. $ u'(x) = \frac{d}{dx}(-\frac{x(4-x)}{6}) $ and $ v'(x) = \frac{d}{dx}(e^x) $.
3. $ u'(x) = -\frac{1}{6} (4 - 2x) $ and $ v'(x) = e^x $.
4. Apply the product rule: $ f'(x) = u'(x)v(x) + u(x)v'(x) $.
5. Substitute the derivatives and the original functions: $ f'(x) = -\frac{1}{6}(4 - 2x)e^x + (-\frac{x(4-x)}{6})e^x $.
6. Simplify the expression: $ f'(x) = -\frac{1}{6}(4e^x - 2xe^x) -\frac{x(4-x)}{6}e^x $.
7. Further simplify: $ f'(x) = -\frac{4e^x}{6} + \frac{2xe^x}{6} -\frac{x(4-x)e^x}{6} $.
8. Combine like terms: $ f'(x) = -\frac{2e^x}{3} + \frac{xe^x}{3} -\frac{x(4-x)e^x}{6} $.

Thus, the derivative of $ f(x) $ is $ f'(x) = -\frac{2e^x}{3} + \frac{xe^x}{3} -\frac{x(4-x)e^x}{6} $.