Product and Chain Rule

[scottcockrum]

Thanks in advance to anyone who helps me with this problem.

I have been out of college a while and am taking a refresher. I have stumbled on this problem which, I think, has a higher level of difficulty and ask for assistance.

Here is the problem.

Find the first derivative of:

(x^2+3)^5(x^2+1/x)^6

Now here is what I think. I initially though this is a simple application of the Product Rule. But I also realized that to find the derivative of each individual term the Chain Rule would be required. I researched my old college text and there are only two examples of the chain rule. The problem being I am not clear on how to differentiate each term in respect to different variables.

Could somebody please provide a solution.

Cheers

Scott C

 

[BigGlenntheHeavy]

\(\displaystyle A \ good \ algebra \ exercise, \ one \ possibility, \ to \ wit;\)

\(\displaystyle f'(x) \ = \ \frac{2(x^{2}+3)^{4}(x^{3}+1)^{5}(11x^{5}+18x^{3}+2x^{2}-9)}{x^{7}}\)

 

[scottcockrum]

If that is the solution, I thank you. Was I correct in thinking that the Product Rule in addition to the Chain Rule is required to solve? Or am I totally off base.

The answer is great...but I need to connect the dots.

 

[BigGlenntheHeavy]

\(\displaystyle Yes, \ I \ used \ both, \ to \ wit:\)

\(\displaystyle f(x) \ = \ (x^{2}+3)^{5}(x^{2}+1/x)^{6} \ = \ \frac{(x^{2}+3)^{5}(x^{3}+1)^{6}}{x^{6}}\)

\(\displaystyle f'(x) \ = \ \frac{x^{6}[5(x^{2}+3)^{4}(2x)(x^{3}+1)^{6}+(x^{2}+3)^{5}(6)(x^{3}+1)^{5}(3x^{2})]-(x^{2}+3)^{5}(x^{3}+1)^{6}(6x^{5})}{x^{12}}\)

\(\displaystyle f'(x) \ = \ \frac{x[10x(x^{2}+3)^{4}(x^{3}+1)^{6}+18x^{2}(x^{2}+3)^{5}(x^{3}+1)^{5}]-6(x^{2}+3)^{5}(x^{3}+1)^{6}}{x^{7}}\)

\(\displaystyle f'(x) \ = \ \frac{10x^{2}(x^{2}+3)^{4}(x^{3}+1)^{6}+18x^{3}(x^{2}+3)^{5}(x^{3}+1)^{5}-6(x^{2}+3)^{5}(x^{3}+1)^{6}}{x^{7}}\)

\(\displaystyle f'(x) \ = \ \frac{(x^{2}+3)^{4}(x^{3}+1)^{5}[10x^{2}(x^{3}+1)+18x^{3}(x^{2}+3)-6(x^{2}+3)(x^{3}+1)]}{x^{7}}\)

\(\displaystyle f'(x) \ = \ \frac{(x^{2}+3)^{4}(x^{3}+1)^{5}[10x^{5}+10x^{2}+18x^{5}+54x^{3}-6(x^{5}+3x^{3}+x^{2}+3)]}{x^{7}}\)

\(\displaystyle f'(x) \ = \ \frac{(x^{2}+3)^{4}(x^{3}+1)^{5}[10x^{5}+10x^{2}+18x^{5}+54x^{3}-6x^{5}-18x^{3}-6x^{2}-18]}{x^{7}}\)

\(\displaystyle f'(x) \ = \ \frac{(x^{2}+3)^{4}(x^{3}+1)^{5}[22x^{5}+36x^{3}+4x^{2}-18]}{x^{7}}\)

\(\displaystyle Finally, \ f'(x) \ = \ \frac{2(x^{2}+3)^{4}(x^{3}+1)^{5}[11x^{5}+18x^{3}+2x^{2}-9]}{x^{7}}, \ I \ quit \ here.\)

 

[scottcockrum]

Thanks. I had forgotten the trick on how to convert the term into something workable you used in step number one.

This might be clear to you, but its not to me so here goes:

Instead of utilizing your trick in step one which then follows with the quotient rule, product rule and then chain rule...wouldn't it be easier to deal with a negative exponent (1/x is equal to x^-1) or quite possibly does the differentiation rules not work with negative exponential powers?

 

[BigGlenntheHeavy]

\(\displaystyle Afterthought:\)

\(\displaystyle Your \ quote: \\)Thanks. I had forgotten the trick on how to convert the term into something workable you used in step number one.

This might be clear to you, but its not to me so here goes:

\(\displaystyle If \ combining \ two \ terms \ is \ not \ clear \ to \ you \ (a \ trick \ as \ you \ so \ aptly \ put \ it), \ then \ I \ suggest \\)

\(\displaystyle you \ refresh \ yourself \ with \ basic \ arithmetic \ instead \ of \ The \ Calculus.\)

\(\displaystyle Note: \ 1+\frac{1}{2} \ = \ \frac{2+1}{2} \ = \ \frac {3}{2}\)