How to find $h'(x)$ given the following?

Question

I need some help with this since I can't seem to figure it out:

$h(x)= f(2x - g(x))$

How do I simplify this in such a way that if I have several values for $f(x)$, $g(x)$, $f'(x)$ and $g'(x)$ and $x$, I can solve for $h'(x)$

x  f(x) g(x) f'(x) g'(x)

-1 -1    0    2    -2
 0  0    1    5     4
 1  4    2    3     1

Find $h'(1)$

Thanks

Are you looking for a general expression $h'(x)$ that you can plug any $x$ into? Or simply $h'(x)$ for a specific value of $x$? — Kyle, Jun 12 '14 at 18:27
And you should include in the post any values for $f(x)$, $g(x)$, $f'(x)$, and $g'(x)$ that you have been provided. — Kyle, Jun 12 '14 at 18:28
A general expression. I have a table which have some values for x and the resulting value for f(x), g(x),f'(x) and g'(x). — Sab ಠ_ಠ, Jun 12 '14 at 18:28
With that information alone, you will only be able to find $h'(x)$ for a handful of values of $x$. — Kyle, Jun 12 '14 at 18:29
Can you post what you get as a general expression for $h'(x)$ using the chain rule? — Tom, Jun 12 '14 at 18:29
I would assume the first row of numbers should be for $x=-1$ rather than $x=1$? — Hayden, Jun 12 '14 at 18:42
@Sabಠ_ಠ Because otherwise you don't have enough information. ;) — Hayden, Jun 12 '14 at 18:46

Hayden · Accepted Answer · 2014-06-12T18:41:48.217

2

The key here is to use the chain rule: $$h'(x)=\frac{d}{dx}(2x-g(x)) \cdot f'(2x-g(x))=(2-g'(x))f'(2x-g(x))$$

If we wish to find $h'(1)$, we start by plugging in $x=1$ to get $$h'(1)=(2-g'(1))f'(2-g(1))=(2-1)f'(2-2)=f'(0)=5$$

edited Jun 12 '14 at 18:41

answered Jun 12 '14 at 18:29

Hayden

I see. Thanks. But let's say I wanted to find $f'(2x-g(x))$, is there a way to simplify it further? – Sab ಠ_ಠ Jun 12 '14 at 18:44
Sadly, without knowing anything more about $f$ and $g$, there's not much that can be said about $f'(2x-g(x))$. – Hayden Jun 12 '14 at 18:46
Okay. I got it now. One last question. How do you know chain rule should be used here and not something else? – Sab ಠ_ಠ Jun 12 '14 at 18:47
Pretty much whenever you see a composition of functions, it's a good indication that chain rule should be used. In this case, we have $f$ composed with the function $2x-g(x)$. – Hayden Jun 12 '14 at 18:48
Aha. Perfect. Thanks a lot :) – Sab ಠ_ಠ Jun 12 '14 at 18:49
Not a problem, let me know if anything isn't clear – Hayden Jun 12 '14 at 18:50
Let us continue this discussion in chat. – Sab ಠ_ಠ Jun 12 '14 at 18:53

1 Answers1