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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!

  • 0
    The ' is for taking the derivative. I was hoping #2 is true regardless of functions d,h.2012-01-26

2 Answers 2

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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.