leading to

If the expression for

`g`(`x`) is first mulitplied out, showing that`g`(`x`) =`x`^{6}-1, the same result follows immediately.-
Let
`k`(`x`) =`g`(`x`)`h`(`x`), so thatThen, using the product rule again, and supressing the dependence on the variable

`x`,and substitution for

`k`(`x`) in terms of`g`(`x`) and`h`(`x`) gives the result.Letting

`g`(`x`) =`h`(`x`) =`f`(`x`) leads to the corollary - Given:
The result of the above exercise gives

and it is quite questionable whether the last expansion is worth the effort; in fact, expanding the original function and then differentiating should lead to the same result. You are encouraged to check this.

watko@mit.edu Last modified August 28, 1998