Let f(x)=(x^2-10x)^4, then the Power Rule and Chain Rule give a derivative of
f'(x)=4(x^2-10x)^3*(2x-10)
=8x^3(x-10)^3(x-5)
f'(x) is therefore clearly zero at x=0,5,10, and these are the critical points of f.
A bit less clearly, f' changes sign from negative to positive as x increases through x=0, from positive to negative as x increases through x=5, and from negative to positive as x increases through x=10. You can plug numbers less than 0, between 0 and 5, between 5 and 10, and greater than 10 and use the continuity of f' to check this. You can also graph f' to confirm it graphically.
The First Derivative Test then implies that the function has local (and global) minima at x=0 and x=10 and a local maximum at x=5.
Interestingly, application of the Second Derivative Test in this example only allows us to conclude where the local maximum is, not where the local minima are (it turns out that f''(0)=f''(10)=0 and f''(5)=-125000<0). In other words, the First Derivative Test is "stronger" (it can be used in more situations) than the Second Derivative Test.
