# Thread: Chain Rule - Proof Using...

1. ## Chain Rule - Proof Using...

I have the question, and part of my work... I can't figure out where to go from here. Bear in mind, this is university level calculus, grade 12... We don't do this type of question too often, so I might need the extra step included in between your regular steps.

Givens:
y=f(x)
f(a)=a
f '(a)=b
g= f o f o f o f o f o f o f o f o f (if I'm using the wrong symbol here, I mean to say f at f at f at f at ....... and so on, 10 times)

Question: find g '(a)

Work/Answer:
g(x)=(f(x))^10
g(a)=(f(a))^10

2. Originally Posted by mike_302
I have the question, and part of my work... I can't figure out where to go from here. Bear in mind, this is university level calculus, grade 12... We don't do this type of question too often, so I might need the extra step included in between your regular steps.

Givens:
y=f(x)
f(a)=a
f '(a)=b
g= f o f o f o f o f o f o f o f o f (if I'm using the wrong symbol here, I mean to say f at f at f at f at ....... and so on, 10 times)

Question: find g '(a)

Work/Answer:
g(x)=(f(x))^10
g(a)=(f(a))^10
?? The problem asked you to find g'(a). Why did you give g(a)?

Try a simple case first. Suppose g(x)= f(f(x)). Then g'(x)= f'(f(x))f'(x). Then g'(a)= f'(f(a))f'(a)= f'(a)f'(a)= $b^2$.

If g(x)= f(f(f(x))), then g'= f'(f(f(x))(f(f(x))' so $g'(a)= f'(f(f(a))(b^2)= f'(a)(b^2)= b^3$.

Now, make a wild guess at what your answer should be!

3. Originally Posted by mike_302
I have the question, and part of my work... I can't figure out where to go from here. Bear in mind, this is university level calculus, grade 12... We don't do this type of question too often, so I might need the extra step included in between your regular steps.

Givens:
y=f(x)
f(a)=a
f '(a)=b
g= f o f o f o f o f o f o f o f o f (if I'm using the wrong symbol here, I mean to say f at f at f at f at ....... and so on, 10 times)

Question: find g '(a)

Work/Answer:
g(x)=(f(x))^10
g(a)=(f(a))^10
Define $f_n(x) = \underbrace{f \circ f \circ \cdots \circ f}_{n}(x)$.

Observe a couple of facts:
1. $f_{n}(x) = f \circ f_{n-1}(x)$ for $n > 1$, and
2. $f_n(a) = a$ for all $n$.

So $g(x) = f_n(x)$.

What is $f_n'(x)$?
Well it is $\left(f \circ f_{n-1}\right)'(x)$. By the chain rule, this is $f' \circ f_{n-1}(x) \cdot f_{n-1}'(x)$.

We continue this process until the subscript is 1 in the rightmost term.

$f_n'(x) = f' \circ f_{n-1}(x) \cdot f' \circ f_{n-2}(x) \cdots f' \circ f_{2}(x) \cdot f' \circ f_1(x) \cdot f_1'(x)$.

By observation 2 above, then:
$f_n'(a) = \underbrace{f'(a) \cdot f'(a) \cdots f'(a) \cdot f'(a)}_{n}$.

Can you take it from there?

4. Ahhh! Clever. I did not think to start off with only the pair and go from there. I was trying to do it with all ten in there.

Okay, that's an interesting way of tackling it.

Appreciated!