Posted by Joel on October 16, 2002 at 01:09:37:
In Reply to: derivatives with exponents posted by Warren Lee on October 15, 2002 at 22:09:19:
: How do you find the derivativeof ((3-x)^(1/3)-x^2)^(1/2). Do I solve (3-x)^(1/3) first and the everything within the original brackets? and then finaly the last exponent?
: Thank you
Work from the outside in, using the chain rule. First look at the whole thing as u^(1/2). The derivative of that is [(1/2)u^(-1/2)]*du/dx. Now, what is du/dx? u is ((3-x)^(1/3)-x^2). So du/dx is the sum of two components: the derivative of -x^2 which is easy (-2x), and the derivative of another composite expression (3-x)^(1/3). To do that one, let v = (3-x) so you have to differentiate v^(1/3) which gives you [(1/3)v^(-2/3)]*dv/dx
v is 3-x, so dv/dx is just -1.
So, now put it all together, replacing u and v with the original expressions, and you get
[(1/2)((3-x)^(1/3)-x^2)^(-1/2)]*[(-1/3)(3-x)^(-2/3) - 2x]