How do you find the first, second derivative of #3x^(2/3)-x^2#?

1 Answer
May 22, 2015

Use the power rule along with "linearity":

#f'(x)=2x^{-1/3}-2x# and #f''(x)=-2/3 x^{-4/3}-2#.

The power rule says that #d/dx(x^{n})=nx^{n-1}# for any fixed number #n#. Linearity of the derivative operator says that #d/dx(af(x)+bg(x))=af'(x)+bg'(x)# for any fixed numbers #a# and #b# and any differentiable functions #f# and #g#.

Linearity can be extended, by using mathematical induction, to any finite "linear combination" of differentiable functions:

#d/dx(a_{1}f_{1}(x)+a_{2}f_{2}(x)+\cdots+a_{n}f_{n}(x))=a_{1}f_{1}'(x)+a_{2}f_{2}'(x)+\cdots+a_{n}f_{n}'(x)#

This can also be written using summation (sigma) notation as:

#d/dx(\sum_{k=1}^{n}a_{k}f_{k}(x))=\sum_{k=1}^{n}a_{k}f_{k}'(x)#

which can be thought of as a "distributive-like" property (even though the symbol #d/dx# is not a number that is being multiplied).

Furthermore, this is often extended to infinite series (sums), especially "power series", as long as #x# is within the interior of the "interval of convergence":

#d/dx(\sum_{k=0}^{\infty}a_{k}(x-c)^{k})=\sum_{k=1}^{\infty}a_{k}k(x-c)^{k-1}# (there's no typo here; the #k=0# case on the right can be omitted)