How do you use the definition of a derivative to find the derivative of #f(x)=(x-6)^(2/3)#, at c=6?

Question

Ezra Adams · Accepted Answer

#f'(6)# does not exist. For how to use the definition, see below. (Warning: It takes a lot of algebra to use the definition to get #f'(x)# for this function.)

For #f(x) = (x-6)^(2/3)#, we need some algebra to use the definition of derivative to find #f'(x)# #f'(x) = lim_(hrarr0) (f(x+h)-f(x))/h# # = lim_(hrarr0) ((x+h-6)^(2/3)-(x-6)^(2/3))/h# # = lim_(hrarr0) (root(3)((x+h-6)^2)-root(3)((x-6)^2))/h#. Obviously (and as expected) this limit has initial form #0/0#. We will write the difference quotient so that the numerator contains no radicals. (Often called "rationalizing" the numerator.) Our goal is to get rid of the #h# in the denominator. Algebra needed The sum of two cubes can be factored: #u^3+v^3 = (u+v)(u^2-uv+v^2)#. (If you didn't learn this in algebra class, learn it now.) In this question, we have #3^"rd"# roots, so we need: #a+b = (root3a+root3b)(root3a^2-root3a root3b+root3b^2)# #= (root3a+root3b)(root3a^2-root3(ab)+root3b^2)# To make matters more interesting, we actually have #3^"rd"# roots of squares. So we need: #a^2+b^2 = (root3(a^2)+root3(b^2))(root3(a^2)^2-root3(a^2) root3(b^2)+root3(b^2)^2)# #= (root3(a^2)+root3(b^2))(root3(a^2)^2-root3(a^2b^2)+root3(b^2)^2)# We also will use #((x+h)-6)^2 = (x+h)^2 - 12(x+h)+36# # = x^2+2xh+h^2-12x-12h+36# And, of course #(x-6)^2 = x^2-12x+36# For the derivative We'll use the algebra discussed above to rewrite the difference quotient. We have #a=(x+h-6)# and #b=(x-6)#. We write: #(root(3)((x+h-6)^2)-root(3)((x-6)^2))/h# # = ((root(3)((x+h-6)^2)-root(3)((x-6)^2)))/h*((root(3)((x+h-6)^2)^2 -root3((x+h-6)^2(x-6)^2)+root(3)((x-6)^2)^2))/((root(3)((x+h-6)^2)^2 -root3((x+h-6)^2(x-6)^2)+root(3)((x-6)^2)^2))# # = ((x+h-6)^2-(x+6)^2)/(h(root(3)((x+h-6)^2)^2 -root3((x+h-6)^2(x-6)^2)+root(3)((x-6)^2)^2))# # = ((x^2+2xh+h^2-12x-12h+36)-(x^2-12x+36))/(h(root(3)((x+h-6)^2)^2 -root3((x+h-6)^2(x-6)^2)+root(3)((x-6)^2)^2))# # = ((x^2+2xh+h^2-12x-12h+36)-(x^2-12x+36))/(h(root(3)((x+h-6)^2)^2 -root3((x+h-6)^2(x-6)^2)+root(3)((x-6)^2)^2))# # = (2xh+h^2-12h)/(h(root(3)((x+h-6)^2)^2 -root3((x+h-6)^2(x-6)^2)+root(3)((x-6)^2)^2))# # = (cancel(h)(2x+h-12))/(cancel(h)(root(3)((x+h-6)^2)^2 -root3((x+h-6)^2(x-6)^2)+root(3)((x-6)^2)^2))# Now we can evaluate the limit as #hrarr0# to get #f'(x) = lim_(hrarr0)((2x+h-12))/((root(3)((x+h-6)^2)^2 -root3((x+h-6)^2(x-6)^2)+root(3)((x-6)^2)^2))# # = (2x-12)/((root(3)((x-6)^2)^2 -root3((x-6)^2(x-6)^2)+root(3)((x-6)^2)^2))# # = (2x-12)/(root(3)(x-6)^4 -root3(x-6)^4+root(3)((x-6)^4))# # = (2(x-6))/(3root(3)(x-6)^4) = (2(x-6))/(3(x-6)root(3)(x-6)) = 2/(3root3(x-6))# At #x=6#, the derivative #f'(x)# does not exist.

Emilia Aguilar · Answer

undefined To find the derivative of $ f(x) = (x - 6)^{\frac{2}{3}} $ at $ c = 6 $, you can use the definition of a derivative. The definition states:

$$ f'(c) = \lim_{h \to 0} \frac{f(c + h) - f(c)}{h} $$

Apply this definition to the given function at $ c = 6 $.