How do you differentiate #f(x)=(x^3)(ln3x)(e^x) # using the product rule?

1 Answer
Oct 11, 2017

Answer:

#f'(x)=(x^3)(ln 3x)(e^x) + (x^2)(e^x) + (3x^2)(e^x)(ln 3x)#

Explanation:

One way to approach this problem is to combine the first two terms (or the last two, your choice) mentally into a single "large" term, and use the Product Rule as you'd expect. When you reach the point of having to take the derivative of the "large" term, you would have to apply the Product Rule to that term off to the side, then substitute the result back in.

For instance:

#f(x) = [(x^3)(ln 3x)]*(e^x) #

#f'(x) = [(x^3)(ln 3x)] * (e^x) + (e^x)*[ (x^3)(3/(3x)) + (ln 3x)(3x^2)] #

#=(x^3)(ln 3x)(e^x) + ((x^3)(e^x))/(x) + (3x^2)(e^x)(ln 3x)#

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

Special note:

The Product Rule can be extended to as many terms as you wish. Simply move one term at a time and take its derivative while keeping the remaining terms intact as-is. You'll note that's essentially what I did above in longhand format. We had 3 "pieces" in the original #f(x)#: #x^3#, #ln 3x#, and #e^x#. Examine closely the three terms in the intermediate derivative work and you can see how two of the three items in each term are original pieces from #f(x)# while the remaining item is a derivative.

Symbolically, this is what I'm noting:

If #f(x) = u*v*w#, with #u, v, w# all functions of #x#, then:

#f'(x) = u'vw + uv'w + uvw'#

If there were 4 functions multiplied together, it would still hold:

#f(x) = u*v*w*z#

#f'(x) = u'vwz + uv'wz + uvw'z + uvwz'#

Why?

#f'(x) = uvwz' + uvw'z + uv'wz + u'vwz #

# = uv(underbrace(wz' + w'z)_ {*(wz)'}) + wz(underbrace(uv' + u'v)_ {(uv)'}) #

# = uv(wz)' + wz*(uv)' #

# = (uv * wz)' = d/(dx) uvwz#