How Do You Calculate the Derivative of Composite Functions Using the Chain Rule?

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1. Find the value of (f o g)^1 at the given value of x.
f(u) = 1-(1/u)
u = g(x) = 1/(1-x)
x = -1




2. Chain rule...



3. Okay, so the derivitive of 1-(1/(1/(1-x))) is 1. Also, the derivitive of 1/(1-x) = 1/(x-1)^2. So, in theory, shouldn't the answer be 1/4? I've solved this in many different ways and I keep getting 1/4 as my answer. However, our answer key says that the answer is 1. Any help would be appreciated.
 
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basenne said:
1. Find the value of (f o g)^1 at the given value of x.
What does (f o g)^1 mean?
basenne said:
f(u) = 1-(1/u)
u = g(x) = 1/(1-x)
x = -1




2. Chain rule...



3. Okay, so the derivitive of 1-(1/(1/(1-x))) is 1. Also, the derivitive of 1/(1-x) = 1/(x-1)^2. So, in theory, shouldn't the answer be 1/4? I've solved this in many different ways and I keep getting 1/4 as my answer. However, our answer key says that the answer is 1. Any help would be appreciated.
 
sorry, that's supposed to be (f of g), meaning f(g(x))
 
Last edited:
By (f of g)^1 do you mean f(g(x))^{-1}?
 
I believe I mean f(g(x))^{1}
 
What exactly does the power of 1 do in f(g(x))?
 
I believe that's supposed to be roman numeral one?

Clearly I have no idea what I'm talking about, oh well, I'll have to ask about it tomorrow.
 
Just solve it with ignoring the exponent because clearly \frac{df(g(x))^1}{dx}= f'(g(x)).g'(x)
 
that's exactly how I solved it, however, I got a different answer than the answer key.
 
  • #10
It's in your calculation then, the exponent doesn't effect it. Nevermind, I think \frac{1}{(x-1)^{-1}} reduces to x leaving your derivative as 1.
 
Last edited:
  • #11
Kevin_Axion said:
It's in your calculation then, the exponent doesn't effect it. Nevermind, I think \frac{1}{(x-1)^{-1}} reduces to x leaving your derivative as 1.
Why do you think that 1/(x - 1)-1 reduces to x?
 
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