
Linda B. answered 09/23/16
Tutor
4.9
(9)
Retired High School Math Teacher
The thing that makes this problem painful is that there is a lot going on. The thing that makes it not so bad, is doing one small step at a time.
You have a quotient, so you will need quotient rule.
You have a radical, and you can use laws of exponents to make that not so bad.
Let's start with the quotient rule:
The quotient rule for f(x)=g(x)/h(x)
f'(x)= (g'(x)h(x)-h'(x)g(x))/[h(x)]2 It looks pretty ugly without math type. I am going to use that old third grade symbol for division to make the numerator more distinct from the denominator.
f'(x)= (g'(x)h(x)-h'(x)g(x))/[h(x)]2 It looks pretty ugly without math type. I am going to use that old third grade symbol for division to make the numerator more distinct from the denominator.
f'(x)= (g'(x)h(x)-h'(x)g(x)) ÷ [h(x)]2 There less ugly.
Let g(x) = √(36-x2) that is the same as (36-x2)(1/2) But the second form is easier to visualize the use of chain rule.
g'(x) = (1/2)(36-x2)(-1/2)(-2x) or -2x/2√(36-x2) = x/√(36-x2)
Let h(x) = 9x, then h'(x)=9 , and [h(x)]2=81x2
Now that you have the parts hashed out. Substitute them carefully, yet fearlessly into the quotient rule. Since you a compound rational, feel free to use the ÷ and ignore the denominator for a bit. It will wait patiently!
Given:
f'(x)= (g'(x)h(x)-h'(x)g(x)) ÷ [h(x)]2
=[ -x·/√(36-x2) · 9x - 9 √(36-x2)]÷ 81x2 Substitution
Factor out and cancel the 9's
We need a LCD, the least common denominator is √(36-x2)
We need a LCD, the least common denominator is √(36-x2)
f'(x)=[ -x2/ √(36-x2) - √(36-x2) · √(36-x2)/√(36-x2)]÷ 9x2
= -x2- (36-x2) / √(36-x2) = (-x2 - 36 + x2)/(9x^2√(36-x^2) ) = -36/ 9x2√(36-x2)
= -4 / x2√(36-x2)
so f'(x) = -4/(x2√(36-x2), just substitute 2 in to find f'(2). And you will be done.
I hope that was helpful.
Contact me if you need more help.