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

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SUMMARY

The discussion focuses on calculating the composite derivative (f o g)' at specific values of x using the chain rule. For the functions f(u) = u^5 + 1 and u = g(x) = sqrt(x) at x = 1, the correct derivative calculation yields (f o g)' = 5(1)^4 * (1/2) = 5/2. The confusion arises from incorrectly substituting u' for u, as u equals 1 when x = 1, not 1/2. The second example with f(u) = 2u/(u^2 + 1) and u = g(x) = 10x^2 + x + 1 at x = 0 confirms the correct application of the chain rule, resulting in f'(1) = 0.

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kuahji
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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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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.
 
That makes sense, thanks for the assistance.
 

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