Lets explore what Wolfram|Alpha can do with more complicated derivatives.

Function composition is when one function is evaluated by the output of another function. For example, \( f( x+h ) \) can be viewed as a composite function where the outside function is \( f(x) \) and the inside function is \( g(x)=x+h \). Notice \( f(g(x)) = f(x+h) \)

Thats fun and all, but the trick for us will be to go backwards. Can you think of two functions that might be hiding in the expression, \( (x^3-1)^2 \)?

For each function below, find \( a(b) \) and \( b(x) \) so that \( f(x) = a( b(x)) \). Also, this is calculus class so might as well compute each derivative!

1. \( f(x) = (x^2+x+1)^3 \)

\( a(b) \) | \( a'(b) \) |

\( b(x) \) | \( b'(x) \) |

2. \( f(x) = \sqrt{ 3x^3 -1 } \)

3. \( f(x) = e^{x^3} \)

4. \( f(x) = \sin(2x^3+x) \)

Compute each \( f'(x) \) above using Wolfram|Alpha. How is each \(f'(x)\) is related to \( a(b), b(x), a'(b), b'(x) \)?