How do you differentiate #f(x)=(x^3)(lnx)(e^x) # using the product rule?

Question

Emma Aldridge · Accepted Answer

#f'(x)=x^2e^x(xlnx+3lnx+1)#

The product rule for three functions basically states that the derivative of the product of three functions equals the derivative of one function multiplied by the other two functions, added to the the other two incarnations of this where the other two functions are differentiated in the other function's place.

Mathematically, this can be written as

#(fgh)^'=f^'gh+fg^'h+fgh^'#

So, for #f(x)=(x^3)(lnx)(e^x)#, we see that

#f'(x)=(x^3)^'(lnx)(e^x)+(x^3)(lnx)^'(e^x)+(x^3)(lnx)(e^x)^'#

Note that:

#{((x^3)^'=3x^2),((lnx)^'=1/x),((e^x)^'=e^x):}#

Therefore

#f'(x)=3x^2(lnx)(e^x)+x^3(1/x)(e^x)+x^3(lnx)(e^x)#

#=3x^2lnx(e^x)+x^2e^x+x^3lnx(e^x)#

#=x^2e^x(xlnx+3lnx+1)#

Emily Ammerman · Answer

undefined To differentiate $ f(x) = (x^3)(\ln x)(e^x) $ using the product rule, follow these steps:

1. Identify the functions that are being multiplied together: $ x^3 $, $ \ln x $, and $ e^x $.
2. 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) $.
3. Let $ u(x) = x^3 $ and $ v(x) = \ln x \cdot e^x $.
4. Find the derivatives $ u'(x) $ and $ v'(x) $.
   - $ u'(x) = 3x^2 $
   - $ v'(x) = (\ln x)'(e^x) + \ln x \cdot (e^x) $
   - Using the product rule again for $ (\ln x)' $, we get $ (\ln x)' = \frac{1}{x} $, so $ v'(x) = \left(\frac{1}{x}\right)(e^x) + \ln x \cdot (e^x) $
5. Plug the derivatives and the original functions back into the product rule formula:
   $$ f'(x) = u'(x)v(x) + u(x)v'(x) $$
   $$ = (3x^2)(\ln x \cdot e^x) + (x^3)\left(\frac{1}{x}\right)(e^x) + (x^3)(\ln x \cdot e^x) $$
6. Simplify the expression:
   $$ f'(x) = 3x^2(\ln x \cdot e^x) + x^2e^x + x^3\frac{1}{x}e^x + x^3(\ln x \cdot e^x) $$
   $$ = 3x^2(\ln x \cdot e^x) + x^2e^x + xe^x + x^3(\ln x \cdot e^x) $$
7. Combine like terms if necessary.