derivative of the function
- Thread starter wjunior30
- Start date
skeeter
Elite Member
- Joined
- Dec 15, 2005
- Messages
- 3,216
neither do I ... do you mean
\(\displaystyle f(x) = (3x^2-x)^4 (x^3+4x)^3\)
???
if so, the product and chain rules are in order ...
\(\displaystyle f'(x) = (3x^2-x)^4 \cdot 3(x^3+4x)^2 \cdot (3x^2 + 4) + (x^3+4x)^3 \cdot 4(3x^2-x)^3 \cdot (6x-1)\)
to simplify, factor out common factors from both terms ...
\(\displaystyle f'(x) = (3x^2-x)^3(x^3+4x)^2[3(3x^2-x)(3x^2+4) + 4(x^3+4x)(6x-1)]\)
I'll leave the remaining algebra "clean-up" for you.
\(\displaystyle f(x) = (3x^2-x)^4 (x^3+4x)^3\)
???
if so, the product and chain rules are in order ...
\(\displaystyle f'(x) = (3x^2-x)^4 \cdot 3(x^3+4x)^2 \cdot (3x^2 + 4) + (x^3+4x)^3 \cdot 4(3x^2-x)^3 \cdot (6x-1)\)
to simplify, factor out common factors from both terms ...
\(\displaystyle f'(x) = (3x^2-x)^3(x^3+4x)^2[3(3x^2-x)(3x^2+4) + 4(x^3+4x)(6x-1)]\)
I'll leave the remaining algebra "clean-up" for you.