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This is a two part question, I just want to verify the following:

  1. ($f\circ h$)$\circ$($d\circ g$)$=f\circ g$ where h is the inverse of d.
  2. (($f\circ h$)$\circ$($d\circ g$))'$=$$[$($f\circ h$)'$\circ $($d\circ g$)$]$($d\circ g$)'

Thank you!

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1  
What does the prime represent? Function inverse? Is $h$ still the inverse of $d$ in the second example? –  Chris Taylor Jan 26 '12 at 8:59
    
The ' is for taking the derivative. I was hoping #2 is true regardless of functions d,h. –  Tom Cruise Jan 26 '12 at 9:27

2 Answers 2

The first part follows directly from the fact that function composition is associative. Observing that $(f \circ h) \circ (d \circ g) = f \circ(h \circ d)\circ g $, and recognizing that the middle term is the identity, the result follows.

As for the second, let $(f \circ h) = p $ and $( d \circ g) = q $ and apply the chain rule to $p \circ q$ and substitute back $p$ and $q$.

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For (1), function composition is associative, so

$$(f\circ h)\circ (d\circ g) = f \circ (h\circ d) \circ g = f \circ e \circ g = f \circ g$$

(where $e$ is the identity function, $e(x)=x$ $\forall x$) as you correctly said.

For (2), Furtuon has already confirmed that it is true in the comments, but we may as well check it. For any two differentiable functions we have

$$(f\circ g)' = g' \cdot (f'\circ g)$$

and hence

$$\left( (f\circ h) \circ (d\circ g) \right)' = (d\circ g)' \cdot ((f\circ h)' \circ (d\circ g))$$

as you wrote.

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