- #1
kdinser
- 337
- 2
I'm having some trouble understanding why the chain rule comes into play in related rates problems. I'm able to follow the procedure outlined in the text and come up with the correct answer, but I'm not really comfortable moving on from a section until I understand why it works.
The standard first example of this problem in the 3 books that I have deals with a balloon blowing up at a constant rate and the rate that the radius of the balloon is changing over time as the volume increases at that rate.
I'm fine with everything in this problem except the part where they take the derivative of both sides and apply the chain rule to this formula.
V=(4*pi*r^3)/3
Actually using the chain rule, I'm fine with and I've gone through enough proofs to understand how and why it works.
h(x)=f(g(x))
h'(x)=f'(g(x))g'(x)
What I'm not getting is, how does it work with related rates problems. When you take the derivative of both sides of V=(4*pi*r^3)/3, where is the f(g(x)) that the chain rule is being applied to?
The standard first example of this problem in the 3 books that I have deals with a balloon blowing up at a constant rate and the rate that the radius of the balloon is changing over time as the volume increases at that rate.
I'm fine with everything in this problem except the part where they take the derivative of both sides and apply the chain rule to this formula.
V=(4*pi*r^3)/3
Actually using the chain rule, I'm fine with and I've gone through enough proofs to understand how and why it works.
h(x)=f(g(x))
h'(x)=f'(g(x))g'(x)
What I'm not getting is, how does it work with related rates problems. When you take the derivative of both sides of V=(4*pi*r^3)/3, where is the f(g(x)) that the chain rule is being applied to?