# Chain rule for product of functions

1. Dec 17, 2015

### smath42

Here is a simple question :

let f(g(x)) = h(x)*g(x).

I want to calculate df/dx.

If I use the product rule, I get g(x)h'(x) + h(x)g'x).

Now if I use the composition/chain rule, I get

df/dx = df/dg * dg/dx = h(x) * g'(x) which is different.

I guess my df/dg = h is wrong, but I can't see what it should be.

Any help is welcome:-)

2. Dec 17, 2015

### pwsnafu

I don't want to double up on $x$, so let's use $y$.
What is $f(y)$ in the question? What is it's derivative?

3. Dec 17, 2015

### smath42

that's where I'm confused...
my first thought that in f(g(x)) f(y) is the function that multiplies y by h(x). therefore if I derivate with respect to y, I get h(x). Which apparently is wrong...

no your question replaces x by y, which in fact doesn't really change anything so I'd say
f(y) obviously equals to h(y)g(y) so its derivative is h'(y)g(y)+g'(y)h(y)...

now I'm again confused

4. Dec 17, 2015

### pwsnafu

This is wrong.

5. Dec 17, 2015

### smath42

humm... so you've seen where my problem is... what's f ?

6. Dec 17, 2015

### smath42

are there 2 f ? the one if f(x) = h(x)*g(x) and the one if f(g(x)) ?

7. Dec 17, 2015

### PeroK

Yes, you've got yourself into a terrible tangle here! If you define another function:

$k(x) = f(g(x)) = h(x)g(x)$

You may be able to see the problem. $f(x) \ne k(x)$

8. Dec 17, 2015

### smath42

OK I see that now, but I'm still in trouble to calculate df/dg. I'd like to say f(u) = h(x)*u but then again I end up with h(x) for df/dg

9. Dec 17, 2015

### pwsnafu

Instead of trying $k(x) = (f \circ g)(x) = h(x) \cdot g(x)$ start with something simpler $k(x) = (f \circ g)(x) = (l \circ g) (x) \cdot g(x)$

10. Dec 17, 2015

### smath42

hum...
then here too, I can't find what l is although I see l(g(x)) should be h(x)

11. Dec 17, 2015

### smath42

Could you guys explain me what is f here please.
I don't know why but I really have trouble seing where is the composition here

12. Dec 17, 2015

### PeroK

It's easier if you set $y = g(x)$, so that $x = g^{-1}(y)$:

$f(y) = f(g(x)) = h(x)g(x) = h(g^{-1}(y))y$

13. Dec 17, 2015

### smath42

hum ok I can now do

f'(y) = h(g^{-1}(y)) + yg^{-1}'(y)h'(g^{-1}(y)) = h(x) + yh'(g^{-1}(y))/g'(g^{-1}(y))
then g'df/dy = g'h+hg'

ok... thanks !

14. Dec 17, 2015

### PeroK

You could have got that directly from the initial equation:

$f(g(x)) = h(x)g(x) \ \Rightarrow \ f'(g(x))g'(x) = h'(x)g(x) + h(x)g'(x)$

15. Dec 17, 2015

### smath42

I'm sorry I must have a very slow mind today but I don't see exactly why the implication and the second equality.

16. Dec 17, 2015

### PeroK

Simply differentiate both sides wrt $x$.

17. Dec 17, 2015

### smath42

ah ok :-D
I was trying to see how one could find h'g+hg' from f'(g)*g' "only"