- #1

- 278

- 0

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?

- Thread starter V0ODO0CH1LD
- Start date

- #1

- 278

- 0

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?

- #2

jbriggs444

Science Advisor

Homework Helper

2019 Award

- 9,336

- 4,032

Perhaps.

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?

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

Science Advisor

- 7,877

- 453

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

- 278

- 0

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

Science Advisor

Homework Helper

2019 Award

- 9,336

- 4,032

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.

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..

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

- 278

- 0

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

- 34,856

- 6,125

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.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?

I assume you mean h(x) = xIf I use mathman's definition I get that d/dx (x^2)^2 = 4x^2

g'(u) = 2u = 2h(x) = 2x

No. jrbriggs444's first post worked this through and got ab. How are you getting (band if I use jrbriggs444's definition I get that d/dx a(bx) = (b^2)ax.

- #8

- 278

- 0

Thanks, haruspex's post actually cleared my confusion!

- Last Post

- Replies
- 2

- Views
- 1K

- Last Post

- Replies
- 2

- Views
- 1K

- Last Post

- Replies
- 6

- Views
- 2K

- Last Post

- Replies
- 4

- Views
- 2K

- Last Post

- Replies
- 5

- Views
- 2K

- Replies
- 1

- Views
- 4K

- Replies
- 4

- Views
- 7K

- Replies
- 2

- Views
- 915

- Last Post

- Replies
- 4

- Views
- 8K

- Last Post

- Replies
- 65

- Views
- 8K