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

[basenne]

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 derivative of 1-(1/(1/(1-x))) is 1. Also, the derivative 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.

 

[Mark44]

1. Find the value of (f o g)^1 at the given value of x.

What does (f o g)^1 mean?

f(u) = 1-(1/u)
u = g(x) = 1/(1-x)
x = -1




2. Chain rule...



3. Okay, so the derivative of 1-(1/(1/(1-x))) is 1. Also, the derivative 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.

 

[basenne]

sorry, that's supposed to be (f of g), meaning f(g(x))

 

[Kevin_Axion]

By (f of g)^1 do you mean f(g(x))^{-1}?

 

[basenne]

I believe I mean f(g(x))^{1}

 

[Kevin_Axion]

What exactly does the power of 1 do in f(g(x))?

 

[basenne]

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.

 

[Kevin_Axion]

Just solve it with ignoring the exponent because clearly \frac{df(g(x))^1}{dx}= f'(g(x)).g'(x)

 

[basenne]

that's exactly how I solved it, however, I got a different answer than the answer key.

 

[Kevin_Axion]

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.

 

[Mark44]

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?