About the chain rule what's wrong with me?

In summary: For example, let u(x) = 2 for all x. Then the quotient above will always be 0, and we have the same issue as above.In summary, the chain rule states that if y and u are differentiable with respect to x, then dy/dx = (dy/du)*(du/dx). However, this example does not work because u is a constant function, which means that du/dx is 0, and dividing by 0 is undefined. This issue arises because u must be nonconstant in order for the derivative to exist.
  • #1
MHD93
93
0
What I know from the chain rule is that if y and u are differentiable with respect to x then dy/dx = (dy/du)*(du/dx)

Now, why is this example doesn't work:
y = x^2
u = c

then we have dy/dx = (dy/du) * (du/dx) = (dy/du) * 0 = 0 doesn't equal 2x
I want an answer irrelated to the chain rule, not explanations involving limits only.

THANKS
 
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  • #2
Mohammad_93 and welcome to the forums.

For this particular problem, you will note that y is not defined in terms of u in any way that exhibits information about change: that is in terms of u against y, u does not contribute any information about y changes with respect to u or vice versa.

For this particular problem you can not use the chain rule for this reason.

An easy way to understand this is to consider a non-chain rule scenario y = c.

Think of the relationship dx/dy = 1/(dy/dx). We know that dy/dx = 0 because we have a flat function which means that we know how y changes with a specific change in x.

But we can't do the reverse. Given a change in y in this specific circumstance, we can not see how a change in y gives a subsequent change in x which is what the dx/dy is actually telling us.

Your chain rule problem is of a somewhat similar nature in that you are trying to find derivatives that do not make sense which means that you get a non-sensical answer.

You get the same kind of problems when we deal with division by 0. For this reason we do not define what dividing by zero actually is because it leads to all kinds of problems in a similar way to what you are getting with your calculus problem.
 
  • #3
I prefer to write the chain rule in the form ##(f\circ g)'(x)=f'(g(x))g'(x)##, where ##f\circ g## is the composition of f and g, defined by ##f\circ g(x)=f(g(x))## for all x. (The left-hand side is interpreted as ##f\circ g## acting on x). You wrote down two equalities that define two functions. If we call the first one f and the other one g, then ##f\circ g## would satisfy ##f\circ g(x)=f(g(x))=f(c)=c^2##. The derivative of this ##f\circ g## is 0, and the chain rule confirms that: ##(f\circ g)'(x)=f'(g(x))g'(x)=2c·0=0##. f'(g(x)) is =2c in this case, because for all t, f'(t)=2t, and if you insert t=g(x)=c, you get f'(g(x))=2c.
 
  • #4
I suppose because u=c is not a function?
 
  • #5
noobilly said:
I suppose because u=c is not a function?

u = c is a perfectly fine function of x.

It's because there is nu function f, such that f(u) = f(c) = x^2.
 
  • #6
"if y and u are differentiable with respect to x"

The issue is that dy/du does not exist. Look at the definition of the derivative:

lim (y(x) - y(x+h))/(u(x) - u(x+h)) = lim (y(x) - y(x+h))/(c - c),

so we are dividing by zero.

For the theorem to work, u must be nonconstant in some neighborhood of the point where we are taking the derivative. Otherwise we run into this issue above.
 

FAQ: About the chain rule what's wrong with me?

What is the chain rule?

The chain rule is a fundamental concept in calculus that allows us to find the derivative of a composite function. In other words, it helps us to calculate the rate of change of a function that is made up of multiple other functions.

Why is the chain rule important?

The chain rule is important because many real-world problems involve functions that are composed of multiple functions. Without the chain rule, we would not be able to accurately calculate the rate of change of these complex functions, making it a crucial tool in mathematics and science.

How does the chain rule work?

The chain rule can be expressed as (f∘g)′(x) = f′(g(x))⋅g′(x), where f and g are functions and f′ and g′ are their respective derivatives. In simpler terms, it tells us that the derivative of a composite function is equal to the derivative of the outer function evaluated at the inner function, multiplied by the derivative of the inner function.

What are some common mistakes when using the chain rule?

One common mistake when using the chain rule is forgetting to apply the derivative to the inner function. Another mistake is using the power rule or product rule on the entire function instead of separately on the inner and outer functions. It is important to carefully identify the inner and outer functions before applying the chain rule.

How can I practice and improve my understanding of the chain rule?

The best way to practice and improve your understanding of the chain rule is to solve a variety of problems that involve composite functions. You can also watch online tutorials or consult with a tutor to gain a deeper understanding of the concept. Additionally, practicing regularly and seeking help when needed can help you improve your skills in using the chain rule.

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