Chain Rule Paradox or Am I Doing Something Wrong?

  • #1
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If h(x) = ax, g(x) = bx and f(x) = g(h(x)).

Wouldn't h'(x) = a? And g'(x) = b? And f'(x) = ab?

But the chain rule says f'(x) must equal h'(x)g'(h(x)), so that means f'(x) = ab(ax) = (a^2)bx.

Am I missing something obvious?
 

Answers and Replies

  • #2
jbriggs444
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If h(x) = ax, g(x) = bx and f(x) = g(h(x)).

Wouldn't h'(x) = a? And g'(x) = b? And f'(x) = ab?

But the chain rule says f'(x) must equal h'(x)g'(h(x)), so that means f'(x) = ab(ax) = (a^2)bx.

Am I missing something obvious?
Perhaps.

h'(x) = a. Yes. It's a constant function.
g'(x) = b. Yes. It's a constant function.

The chain rule says that f'(x) = h'(x)g'(h(x)). Yes.

What's g'(h(x)) ?

That's g'(whatever) = b

What's h'(x)?

That's h'(whatever) = a

What's h'(x)g'(h(x)) ?

That's ab, just like it is supposed to be.
 
  • #3
mathman
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f(x)=g(h(x))=bh(x). f'(x)=bh'(x)=ba. The point is that g'(u)=b, no matter what u is, even if u=h(x).
 
  • #4
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Okay, but if h(x) = x^2, g(x) = x^2 and f(x) = g(h(x))

Now by your logic f'(x) should be 4x^2 not 4x^3, since h'(x) = 2x and g'(x) = 2x.
Then f'(x) = 2x * 2x = 2x^2.

I am still confused..
 
  • #5
jbriggs444
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Okay, but if h(x) = x^2, g(x) = x^2 and f(x) = g(h(x))

Now by your logic f'(x) should be 4x^2 not 4x^3, since h'(x) = 2x and g'(x) = 2x.
Then f'(x) = 2x * 2x = 2x^2.

I am still confused..
In this case the first derivitives are not constant functions. You need to keep track of what your x's are. Don't let the letters trick you.

Work it through. Chain rule says h'(x)g'(h(x))

h'(x) = 2x.
h(x) = x^2
remember that g'(y) = 2y
so
g'(h(x)) = 2h(x) = 2x^2

Put it together, f'(x) = 2x * 2x^2 = 4x^3

And since f(x) = h(x)^2 = (x^2)^2 = x^4, this result is correct.
 
  • #6
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So if both equations in the chain are linear I use g'(whatever) and h'(whatever), but if one of them isn't what I take the derivative with respect to matters? Why?

If I use mathman's definition I get that d/dx (x^2)^2 = 4x^2 and if I use jrbriggs444's definition I get that d/dx a(bx) = (b^2)ax. What is wrong?
 
  • #7
haruspex
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So if both equations in the chain are linear I use g'(whatever) and h'(whatever), but if one of them isn't what I take the derivative with respect to matters?
No, it's entirely consistent. You just have to understand that g' here does not mean differentiate g wrt x. It means differentiate g wrt whatever g is a function of, namely h.
If I use mathman's definition I get that d/dx (x^2)^2 = 4x^2
I assume you mean h(x) = x2, g(u) = u2, where u = h.
g'(u) = 2u = 2h(x) = 2x2; h'(x) = 2x; chain rule gives 2x2.2x = 4x3
and if I use jrbriggs444's definition I get that d/dx a(bx) = (b^2)ax.
No. jrbriggs444's first post worked this through and got ab. How are you getting (b2)ax?
 
  • #8
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Thanks, haruspex's post actually cleared my confusion!
 

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