How do you differentiate #f(x)= (3x+1)^4(2x-3)^3 # using the product rule?

1 Answer
Sep 30, 2016

Answer:

#12(3x+1)^3(2x-3)^3 + 6(2x-3)^2(3x+1)^4#

Explanation:

This problem actually involves two rules: the product rule, and the chain rule. But first, let's just focus on the product rule. The general equation for this is:

#d/dx[f(x)*g(x)] = d/dx[f(x)]*g(x) + d/dx[g(x)]*f(x)#

Now, we just plug in everything:

#d/dx[(3x+1)^4]##(2x-3)^3# #+ d/dx[(2x-3)^3]##(3x-1)^4#

The problem arises at the fact that your two functions are actually compositions of functions. For example, #(3x+1)^4# is really the composition of the two functions #x^4# and #3x+1#. Hence, we'll need to use the chain rule to evaluate these derivatives. Let's take each one in turn:

#d/dx[(3x+1)^4]#

So first, we do the outermost function (in this case #x^4#)'s derivative, then multiply by the derivative of the inner function (#3x+1#). So, this gives us:

=> #4(3x+1)^3 * d/dx(3x+1)#
=> #4(3x+1)^3 * 3#
=> #12(3x+1)^3#

Now, for the second one:

#d/dx[(2x-3)^3]#

Again, same process. Derivative of the outside, then multiply by derivative of the inside:

=> #3(2x-3)^2 * d/dx(2x-3)#
=> #3(2x-3)^2 * 2#
=> #6(2x-3)^2#

We're done! Now, we just go ahead and plug these back into the product rule equation we got for our final answer:

#12(3x+1)^3(2x-3)^3 + 6(2x-3)^2(3x+1)^4#

You could foil out some stuff here, but in my opinion, it's really just more work. Easier to leave it as is.

Hope that helped :)