Calculus
posted by Lauren on .
I have a two part question that pertains to a curve (r(x)) and its tangent line at x=3.
We are given that at x=3, r(x)=8. In order to find the slope of the tangent line, we are given another point (on the tangent line): (3.2, 8.5). Therefore the slope of the tangent line at x=3 is 5/2
A. Let c(x)= 1/r(x). Find c'(3).
I used the quotient rule (for clarity, I am substituting "a" for r(x))
c'(x)= (a[0]  1a')/a^2
Which reduces to
a'/a^2
Replacing r(x) at x=3 for a, we get:
r'(3)/(r(3))^2
Since the derivative of r(x) at 3 is equal to the slope of the tangent line at x=3, we get
(5/2)/(8)^2 or 5/128
Does that look correct?
b. Let i(x) = the inverse of r(x). Find i'(8).
I had some trouble with this one. I know that the point (3,8) is on r(x) (from the given information in part a). This would mean that if i(x) is the inverse of r(x), then (8,3) would be on i(x). I also know that r'(3)=5/2. Because derivatives of inverses are reciprocals at corresponding points, would this mean that i'(8)=2/5?

I agree with your thinking for both questions.
I especially like that you realize the relationship of slopes between corresponding points on a relation and its inverse.