Find (f o g)' at x=1: Composite Derivative

In summary, the conversation discusses finding the value of the composite function (f o g)' at a given value of x. The process involves finding the derivatives of both f(u) and u=g(x), and then using the chain rule to calculate the final derivative. In the first example, the mistake was made in plugging in the values for u and u' in the wrong places, whereas the second example was solved correctly.
  • #1
kuahji
394
2
Find the value of (f o g)' at the given value of x.

f(u)=u^5+1, u=g(x)=sqrt x, x = 1

So I found the derivate of f(u) & of u.
f'(u)= 5u^4 & u'= 1/(2sqrt x)

Then I plugged 1 in for x in u' & got 1/2. What I don't understand is why I can't just then plug 1/2 in for u & solve.

Another example
f(u)= 2u/(u^2+1), u=g(x)= 10x^2+x+1, x=0

Here I did the same, found the derivative of f'(u) & u'.
f'(u) = -2u^2+2/(u^2+1)^2 & u'=20x+1

Then I plugged in zero for u' & got 1. If you then plug 1 into f'(u), you get 0. Which matches the answer in the back of the book, but the first one does not using the same technique. Somewhere I think my knowledge of the concept is incomplete. Any help would be welcome.
 
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  • #2
kuahji said:
Find the value of (f o g)' at the given value of x.

f(u)=u^5+1, u=g(x)=sqrt x, x = 1

So I found the derivate of f(u) & of u.
f'(u)= 5u^4 & u'= 1/(2sqrt x)

Then I plugged 1 in for x in u' & got 1/2. What I don't understand is why I can't just then plug 1/2 in for u & solve.
Yes, when x= 1, u'= 1/2. One obvious reason why you can't "plug 1/2 in for u & solve" is that u IS not 1/2! When x= 1, u= sqrt(1)= 1. It is u' that is 1/2. By the chain rule, (fo g)'= df/du du/dx= 5(1)^4 (1/2= 5/2.

Another example
f(u)= 2u/(u^2+1), u=g(x)= 10x^2+x+1, x=0

Here I did the same, found the derivative of f'(u) & u'.
f'(u) = -2u^2+2/(u^2+1)^2 & u'=20x+1[/quote[
Better to put in parentheses: f'(u)= (-2u+ 2)/(u^2+1)^2

[qu0te]Then I plugged in zero for u' & got 1. If you then plug 1 into f'(u), you get 0. Which matches the answer in the back of the book, but the first one does not using the same technique. Somewhere I think my knowledge of the concept is incomplete. Any help would be welcome.
Yes, what you got for the second problem, f'(1)= 0 is correct. The problem with the first may have been distinguishing between u and u'. When x= 1, u= 1 and u'= 1/2. Use them in the correct places.
 
  • #3
That makes sense, thanks for the assistance.
 

1. What is the formula for finding the composite derivative at x=1?

The formula for finding the composite derivative at x=1 is (f o g)'(x) = f'(g(x)) * g'(x).

2. How do you find the composite derivative when given two functions, f(x) and g(x)?

To find the composite derivative at x=1, you must first take the derivative of the outer function, f'(g(x)), and then multiply it by the derivative of the inner function, g'(x).

3. Can the order of the functions be switched when finding the composite derivative at x=1?

No, the order of the functions cannot be switched when finding the composite derivative at x=1. The outer function must always be applied first, followed by the inner function.

4. What is the importance of finding the composite derivative at x=1?

Finding the composite derivative at x=1 allows you to easily calculate the instantaneous rate of change of a composite function at a specific point. This can be useful in many applications, such as physics and engineering.

5. Are there any special cases to consider when finding the composite derivative at x=1?

Yes, when the outer function is a constant, the composite derivative at x=1 will simply be the derivative of the inner function, g'(x). Additionally, when the inner function is a constant, the composite derivative at x=1 will be equal to 0.

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