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

• basenne
In summary, the conversation is about finding the value of (f o g)^1 at a given value of x, where f(u) = 1-(1/u) and u = g(x) = 1/(1-x). The discussion also touches on the chain rule and solving for the derivative of 1-(1/(1/(1-x))), which results in a different answer than the given answer key. There is confusion about the notation and the role of the exponent in the calculation.
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 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.

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.

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:
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?

## What are derivatives of functions?

Derivatives of functions are a mathematical concept used to describe the rate of change of a function at a specific point. It represents the slope of the tangent line at that point on the function's graph.

## How do you find the derivative of a function?

The derivative of a function can be found using the rules of differentiation, such as the power rule, product rule, and quotient rule. These rules involve taking the derivative of each term in the function and combining them to find the overall derivative.

## What is the purpose of finding derivatives of functions?

Finding derivatives of functions is important in many areas of science and engineering, as it allows us to understand and analyze the behavior of functions and their rates of change. It is used in fields such as physics, economics, and engineering to solve real-world problems.

## What is the difference between derivatives and differentials?

Derivatives and differentials are closely related concepts, but they are not the same. A derivative represents the rate of change of a function, while a differential represents the change in the function's value. In other words, a derivative is a ratio, while a differential is an actual value.

## Can derivatives be negative?

Yes, derivatives can be negative. This means that the function is decreasing at that point, or that its slope is negative. It is important to consider the sign of the derivative when analyzing the behavior of a function at a specific point.

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