Welcome to our community

Be a part of something great, join today!

hp1357's question at Yahoo! Answers regarding an extended product rule

  • Thread starter
  • Admin
  • #1


Staff member
Feb 24, 2012
Here is the question:

The Product Rule, Calculus Help?

This is a problem from my calculus textbooks:

Determine an expression for f'(x) is f(x)=g1(x)g2(x)g3(x)...gn-1(x), gn(x). If f(x)=(1+x)(1+2x)(1+3x)...(1+nx) find f'(0).
I have posted a link there to this thread so that the OP may view my work.
  • Thread starter
  • Admin
  • #2


Staff member
Feb 24, 2012
Hello hp1357,

a) The product rule for a composite function that is the product of two functions is well-known and will be the basis for working this problem (and accepted without proof):

\(\displaystyle \frac{d}{dx}\left(g_1(x)\cdot g_2(x) \right)=g_1'(x)\cdot g_2(x)+g_1(x)\cdot g_2'(x)\)

Using this rule, let's look at:

\(\displaystyle \frac{d}{dx}\left(g_1(x)\cdot g_2(x)\cdot g_3(x) \right)\)

Now, let's associate two of the functions together, it doesn't matter which two, so let's use the first two:

\(\displaystyle \frac{d}{dx}\left(\left(g_1(x)\cdot g_2(x) \right)\cdot g_3(x) \right)\)

Now, using the product rule above, we may state:

\(\displaystyle \frac{d}{dx}\left(\left(g_1(x)\cdot g_2(x) \right)\cdot g_3(x) \right)=\frac{d}{dx}\left(g_1(x)\cdot g_2(x) \right)\cdot g_3(x)+\left(g_1(x)\cdot g_2(x) \right)\cdot g_3'(x)\)

Using the product rule again, we find:

\(\displaystyle \frac{d}{dx}\left(\left(g_1(x)\cdot g_2(x) \right)\cdot g_3(x) \right)=\left(g_1'(x)\cdot g_2(x)+g_1(x)\cdot g_2'(x) \right)\cdot g_3(x)+\left(g_1(x)\cdot g_2(x) \right)\cdot g_3'(x)\)

And distributing, we find:

\(\displaystyle \frac{d}{dx}\left(\left(g_1(x)\cdot g_2(x) \right)\cdot g_3(x) \right)=g_1'(x)\cdot g_2(x)\cdot g_3(x)+g_1(x)\cdot g_2'(x)\cdot g_3(x)+g_1(x)\cdot g_2(x)\cdot g_3'(x)\)

Now, this is enough to suggest the pattern (our induction hypothesis $P_n$):

\(\displaystyle \frac{d}{dx}\left[\prod_{k=1}^n\left(g_k(x) \right) \right]=\sum_{k=1}^n\left[\prod_{j=1}^{k-1}\left(g_j(x) \right)\cdot\frac{d}{dx}\left(g_k(x) \right)\cdot\prod_{j=k+1}^n\left(g_j(x) \right) \right]\)

Next, consider:

\(\displaystyle \frac{d}{dx}\left[\prod_{k=1}^n\left(g_k(x) \right)\cdot g_{n+1}(x) \right]\)

Using the product rule, and incorporating the new factor into the product. we may state:

\(\displaystyle \frac{d}{dx}\left[\prod_{k=1}^{n+1}\left(g_k(x) \right) \right]=\frac{d}{dx}\left[\prod_{k=1}^n\left(g_k(x) \right) \right]\cdot g_{n+1}(x)+\prod_{k=1}^n\left(g_k(x) \right)\cdot g_{n+1}'(x)\)

Using our induction hypothesis, this becomes:

\(\displaystyle \frac{d}{dx}\left[\prod_{k=1}^{n+1}\left(g_k(x) \right) \right]=\sum_{k=1}^n\left[\prod_{j=1}^{k-1}\left(g_j(x) \right)\cdot\frac{d}{dx}\left(g_k(x) \right)\cdot\prod_{j=k+1}^n\left(g_j(x) \right) \right]\cdot g_{n+1}(x)+\prod_{k=1}^n\left(g_k(x) \right)\cdot g_{n+1}'(x)\)

Now, incorporating the factor at the end of the first term on the right, we have:

\(\displaystyle \frac{d}{dx}\left[\prod_{k=1}^{n+1}\left(g_k(x) \right) \right]=\sum_{k=1}^n\left[\prod_{j=1}^{k-1}\left(g_j(x) \right)\cdot\frac{d}{dx}\left(g_k(x) \right)\cdot\prod_{j=k+1}^{n+1}\left(g_j(x) \right) \right]+\prod_{k=1}^n\left(g_k(x) \right)\cdot g_{n+1}'(x)\)

And finally incorporating the second term on the right within the first summation term, we have:

\(\displaystyle \frac{d}{dx}\left[\prod_{k=1}^{n+1}\left(g_k(x) \right) \right]=\sum_{k=1}^{n+1}\left[\prod_{j=1}^{k-1}\left(g_j(x) \right)\cdot\frac{d}{dx}\left(g_k(x) \right)\cdot\prod_{j=k+1}^{n+1}\left(g_j(x) \right) \right]\)

We have derived $P_{n+1}$ from $P_n$, thereby completing the proof by induction.

b) Now, if:

\(\displaystyle f(x)=\prod_{k=1}^n\left(g_k(x) \right)\)


\(\displaystyle g_k(x)=(1+kx)\), we see that we have:

\(\displaystyle f'(x)=\sum_{k=1}^n\left[\prod_{j=1}^{k-1}\left(1+jx \right)\cdot k\cdot\prod_{j=k+1}^n\left(1+jx \right) \right]\)


\(\displaystyle f'(0)=\sum_{k=1}^n(k)=\frac{n(n+1)}{2}\)