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How to explain what's wrong with this application of the chain rule?

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2

$begingroup$

Yesterday a student in my calculus class attempted something like this:

Task: Derive $3^5x+1$ with respect to $x$.

Proposed solution: Let the inner function be $g(x)=3$ and the outer function be $f(z)=z^5x+1$, so that $f(g(x))=3^5x+1$. Then $f'(z)=(5x+1)cdot z^5x$ and $g'(x)=0$, so by the chain rule
$$
fracd (3^5x+1)dx = f'(g(x))g'(x)=0.
$$

I had difficulties to explain what's wrong with this and basically just told the student "the" right way to do it. Although I now have a rough idea of what's wrong, I'd like to hear from others:

1. Have you seen similar attempts?


2. How would you explain to a beginning calculus student what's wrong with this specific solution?

calculus

share|improve this question

asked 4 hours ago

Michael BächtoldMichael Bächtold

566312

$endgroup$

add a comment | 

2

$begingroup$

Yesterday a student in my calculus class attempted something like this:

Task: Derive $3^5x+1$ with respect to $x$.

Proposed solution: Let the inner function be $g(x)=3$ and the outer function be $f(z)=z^5x+1$, so that $f(g(x))=3^5x+1$. Then $f'(z)=(5x+1)cdot z^5x$ and $g'(x)=0$, so by the chain rule
$$
fracd (3^5x+1)dx = f'(g(x))g'(x)=0.
$$

I had difficulties to explain what's wrong with this and basically just told the student "the" right way to do it. Although I now have a rough idea of what's wrong, I'd like to hear from others:

1. Have you seen similar attempts?


2. How would you explain to a beginning calculus student what's wrong with this specific solution?

calculus

share|improve this question

asked 4 hours ago

Michael BächtoldMichael Bächtold

566312

$endgroup$

add a comment | 

$begingroup$

Yesterday a student in my calculus class attempted something like this:

Task: Derive $3^5x+1$ with respect to $x$.

Proposed solution: Let the inner function be $g(x)=3$ and the outer function be $f(z)=z^5x+1$, so that $f(g(x))=3^5x+1$. Then $f'(z)=(5x+1)cdot z^5x$ and $g'(x)=0$, so by the chain rule
$$
fracd (3^5x+1)dx = f'(g(x))g'(x)=0.
$$

I had difficulties to explain what's wrong with this and basically just told the student "the" right way to do it. Although I now have a rough idea of what's wrong, I'd like to hear from others:

1. Have you seen similar attempts?


2. How would you explain to a beginning calculus student what's wrong with this specific solution?

calculus

share|improve this question

asked 4 hours ago

Michael BächtoldMichael Bächtold

566312

$endgroup$

share|improve this question

asked 4 hours ago

Michael BächtoldMichael Bächtold

566312

share|improve this question

asked 4 hours ago

Michael BächtoldMichael Bächtold

566312

asked 4 hours ago

Michael BächtoldMichael Bächtold

566312

asked 4 hours ago

Michael BächtoldMichael Bächtold

566312

2 Answers
2

active

oldest

votes

3

$begingroup$

f is not a function of (only) z - f here is a function of x as well as z. I think this explanation is intelligible to a calc 1 student, and gets at the heart of the matter.

share|improve this answer

answered 3 hours ago

Henry TowsnerHenry Towsner

6,9652349

$endgroup$

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  Hmm, so the student should reply: as soon as $f(x)$ contains parameters other than $x$ I am not allowed to apply the chain rule?
  $endgroup$
  – Michael Bächtold
  3 hours ago
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  By the way: to my mind $f$ ist not a function of $z$ at all. Maybe you meant $f(z)$?
  $endgroup$
  – Michael Bächtold
  3 hours ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @MichaelBächtold: The student should know that if f(x) contains variables other than x then the chain rule doesn't apply. (This might be an opportunity to mention that there is a variant of the chain rule to be learned later for covering such situations.) I decline to get into a pedantic discussion of the distinction between f and f(z).
  $endgroup$
  – Henry Towsner
  2 hours ago
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  I'm quite sure you use the chain rule to derive things that contain more than just $x$ in your calculus class, like $sqrtx^2+k$. You might say: that's ok if we treat $k$ as a constant and not as a variable. But then student might then ask: why am I not allowed to treat $x$ as a constant in the definition of $f$? (And a mathematician might add: what's the difference between a variable and a constant?). Apologies if my pedantry offends you.
  $endgroup$
  – Michael Bächtold
  1 hour ago
  

  
  
  
  


• 
  
  
  

  
  2
  

  
  
  
  
  
  

  
  $begingroup$
  @MichaelBächtold: In general, the difference between a variable and a constant is contextual and tricky to make precise, but for purposes of the chain rule in this case, x is a variable because we're taking the derivative with respect to it. I find that students don't usually have difficulty with this point (for instance, one could imagine students getting confused about the difference between the derivative of f(x)=c and f(x)=x, but that's not a particularly common issue), because it's a clean syntactic rule and it's backed up by the notion (x and z are conventionally variables, k isn't).
  $endgroup$
  – Henry Towsner
  1 hour ago
  

  
  
  
  

add a comment | 

2

$begingroup$

This is a VERY VERY typical problem. In fact, it's a problem even for $fracddx3^x$, much less your example.

The way I try to deal with this is one of two ways.

1. What has to happen first? To evaluate $3^5x+1$, you have to evaluate $5x+1$ first. So that is the inside function in the chain rule, just like in $sin(x^2)$ you have $x^2$ to evaluate first, so it is the inside function.




2. You could rethink how we notate or talk about exponential functions. In particular, Excel has $e^x$ written as exp(x) (I think as an option). So one can ask what the "input" is here.

However, on the plus side the student does seem to have the chain rule down; it's just the exponential notation that is causing trouble. So there is definitely hope here. And again, you should not be surprised at encountering this, so it is worth your time to come up with several possible responses for it in the long run. Good luck!

share|improve this answer

answered 52 mins ago

kcrismankcrisman

3,480730

$endgroup$

add a comment | 

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3

$begingroup$

f is not a function of (only) z - f here is a function of x as well as z. I think this explanation is intelligible to a calc 1 student, and gets at the heart of the matter.

share|improve this answer

answered 3 hours ago

Henry TowsnerHenry Towsner

6,9652349

$endgroup$

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  Hmm, so the student should reply: as soon as $f(x)$ contains parameters other than $x$ I am not allowed to apply the chain rule?
  $endgroup$
  – Michael Bächtold
  3 hours ago
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  By the way: to my mind $f$ ist not a function of $z$ at all. Maybe you meant $f(z)$?
  $endgroup$
  – Michael Bächtold
  3 hours ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @MichaelBächtold: The student should know that if f(x) contains variables other than x then the chain rule doesn't apply. (This might be an opportunity to mention that there is a variant of the chain rule to be learned later for covering such situations.) I decline to get into a pedantic discussion of the distinction between f and f(z).
  $endgroup$
  – Henry Towsner
  2 hours ago
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  I'm quite sure you use the chain rule to derive things that contain more than just $x$ in your calculus class, like $sqrtx^2+k$. You might say: that's ok if we treat $k$ as a constant and not as a variable. But then student might then ask: why am I not allowed to treat $x$ as a constant in the definition of $f$? (And a mathematician might add: what's the difference between a variable and a constant?). Apologies if my pedantry offends you.
  $endgroup$
  – Michael Bächtold
  1 hour ago
  

  
  
  
  


• 
  
  
  

  
  2
  

  
  
  
  
  
  

  
  $begingroup$
  @MichaelBächtold: In general, the difference between a variable and a constant is contextual and tricky to make precise, but for purposes of the chain rule in this case, x is a variable because we're taking the derivative with respect to it. I find that students don't usually have difficulty with this point (for instance, one could imagine students getting confused about the difference between the derivative of f(x)=c and f(x)=x, but that's not a particularly common issue), because it's a clean syntactic rule and it's backed up by the notion (x and z are conventionally variables, k isn't).
  $endgroup$
  – Henry Towsner
  1 hour ago
  

  
  
  
  

add a comment | 

3

$begingroup$

f is not a function of (only) z - f here is a function of x as well as z. I think this explanation is intelligible to a calc 1 student, and gets at the heart of the matter.

share|improve this answer

answered 3 hours ago

Henry TowsnerHenry Towsner

6,9652349

$endgroup$

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  Hmm, so the student should reply: as soon as $f(x)$ contains parameters other than $x$ I am not allowed to apply the chain rule?
  $endgroup$
  – Michael Bächtold
  3 hours ago
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  By the way: to my mind $f$ ist not a function of $z$ at all. Maybe you meant $f(z)$?
  $endgroup$
  – Michael Bächtold
  3 hours ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @MichaelBächtold: The student should know that if f(x) contains variables other than x then the chain rule doesn't apply. (This might be an opportunity to mention that there is a variant of the chain rule to be learned later for covering such situations.) I decline to get into a pedantic discussion of the distinction between f and f(z).
  $endgroup$
  – Henry Towsner
  2 hours ago
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  I'm quite sure you use the chain rule to derive things that contain more than just $x$ in your calculus class, like $sqrtx^2+k$. You might say: that's ok if we treat $k$ as a constant and not as a variable. But then student might then ask: why am I not allowed to treat $x$ as a constant in the definition of $f$? (And a mathematician might add: what's the difference between a variable and a constant?). Apologies if my pedantry offends you.
  $endgroup$
  – Michael Bächtold
  1 hour ago
  

  
  
  
  


• 
  
  
  

  
  2
  

  
  
  
  
  
  

  
  $begingroup$
  @MichaelBächtold: In general, the difference between a variable and a constant is contextual and tricky to make precise, but for purposes of the chain rule in this case, x is a variable because we're taking the derivative with respect to it. I find that students don't usually have difficulty with this point (for instance, one could imagine students getting confused about the difference between the derivative of f(x)=c and f(x)=x, but that's not a particularly common issue), because it's a clean syntactic rule and it's backed up by the notion (x and z are conventionally variables, k isn't).
  $endgroup$
  – Henry Towsner
  1 hour ago
  

  
  
  
  

add a comment | 

$begingroup$

f is not a function of (only) z - f here is a function of x as well as z. I think this explanation is intelligible to a calc 1 student, and gets at the heart of the matter.

share|improve this answer

answered 3 hours ago

Henry TowsnerHenry Towsner

6,9652349

$endgroup$

share|improve this answer

answered 3 hours ago

Henry TowsnerHenry Towsner

6,9652349

answered 3 hours ago

Henry TowsnerHenry Towsner

6,9652349

answered 3 hours ago

Henry TowsnerHenry Towsner

6,9652349

2

$begingroup$

This is a VERY VERY typical problem. In fact, it's a problem even for $fracddx3^x$, much less your example.

The way I try to deal with this is one of two ways.

1. What has to happen first? To evaluate $3^5x+1$, you have to evaluate $5x+1$ first. So that is the inside function in the chain rule, just like in $sin(x^2)$ you have $x^2$ to evaluate first, so it is the inside function.




2. You could rethink how we notate or talk about exponential functions. In particular, Excel has $e^x$ written as exp(x) (I think as an option). So one can ask what the "input" is here.

However, on the plus side the student does seem to have the chain rule down; it's just the exponential notation that is causing trouble. So there is definitely hope here. And again, you should not be surprised at encountering this, so it is worth your time to come up with several possible responses for it in the long run. Good luck!

share|improve this answer

answered 52 mins ago

kcrismankcrisman

3,480730

$endgroup$

add a comment | 

2

$begingroup$

This is a VERY VERY typical problem. In fact, it's a problem even for $fracddx3^x$, much less your example.

The way I try to deal with this is one of two ways.

1. What has to happen first? To evaluate $3^5x+1$, you have to evaluate $5x+1$ first. So that is the inside function in the chain rule, just like in $sin(x^2)$ you have $x^2$ to evaluate first, so it is the inside function.




2. You could rethink how we notate or talk about exponential functions. In particular, Excel has $e^x$ written as exp(x) (I think as an option). So one can ask what the "input" is here.

However, on the plus side the student does seem to have the chain rule down; it's just the exponential notation that is causing trouble. So there is definitely hope here. And again, you should not be surprised at encountering this, so it is worth your time to come up with several possible responses for it in the long run. Good luck!

share|improve this answer

answered 52 mins ago

kcrismankcrisman

3,480730

$endgroup$

add a comment | 

$begingroup$

This is a VERY VERY typical problem. In fact, it's a problem even for $fracddx3^x$, much less your example.

The way I try to deal with this is one of two ways.

1. What has to happen first? To evaluate $3^5x+1$, you have to evaluate $5x+1$ first. So that is the inside function in the chain rule, just like in $sin(x^2)$ you have $x^2$ to evaluate first, so it is the inside function.




2. You could rethink how we notate or talk about exponential functions. In particular, Excel has $e^x$ written as exp(x) (I think as an option). So one can ask what the "input" is here.

However, on the plus side the student does seem to have the chain rule down; it's just the exponential notation that is causing trouble. So there is definitely hope here. And again, you should not be surprised at encountering this, so it is worth your time to come up with several possible responses for it in the long run. Good luck!

share|improve this answer

answered 52 mins ago

kcrismankcrisman

3,480730

$endgroup$

share|improve this answer

answered 52 mins ago

kcrismankcrisman

3,480730

answered 52 mins ago

kcrismankcrisman

3,480730

answered 52 mins ago

kcrismankcrisman

3,480730