My Math Forum Chain Rule QUestion

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 October 21st, 2009, 01:33 PM #1 Senior Member   Joined: Feb 2008 Posts: 100 Thanks: 0 Chain Rule QUestion I have a question in regards to one of the chain rules for finding the dirivative of a composite functions, my question is on the logarithmic rule that follows, y = log(base7)x y = logx/log7 y = 1/log7 * logx [color=#BF0000]y(primed) = 1/log7 * 1/x[/color] y(primed) = 1/x(log7) Now in red is 1/log7 being treated as a constant or coefficient? Weren't they suppose to us the product rule for finding a derivative?
 October 21st, 2009, 03:13 PM #2 Senior Member   Joined: Jan 2009 From: Japan Posts: 192 Thanks: 0 Re: Chain Rule QUestion The value $\frac{1}{\log 7}$ does not depend on x, so it is treated as a constant coefficient. You can treat it with the product rule without getting the wrong result since $\frac{d}{dx}\left(\frac{1}{\log 7}\right) = 0\\ \frac{d}{dx} \left(\frac{1}{\log 7} * \log x\right) = \underbrace{\frac{d}{dx}\left(\frac{1}{\log 7}\right)}_{\ 0} * \log x + \frac{1}{\log 7} * \underbrace{\frac{d}{dx}\left(\log x\right)}_{\ \frac{1}{x}} = \frac{1}{x*\log 7}$ so the distinction is of interest for practical reasons, not because they'll yield different answers.
 October 21st, 2009, 09:01 PM #3 Senior Member   Joined: Feb 2008 Posts: 100 Thanks: 0 Re: Chain Rule QUestion Hmm that's odd I posted a reply earlier at school and it didn't go through.... Well anywas thanks for clarifying things up cmusik, but another quick questions.... Why is d/dx of [color=#BF0000]e^(-x)[/color] = [color=#BF0000]-e^(-x)[/color]? Shouldn't it just be [color=#BF0000]e^(-x)[/color]?
 October 21st, 2009, 09:09 PM #4 Senior Member   Joined: Jan 2009 From: Japan Posts: 192 Thanks: 0 Re: Chain Rule QUestion The chain rule applies here: $\frac{d}{dx} e^{-x}= e^{-x} \frac{d}{dx} (-x) = e^{-x}\ (-1)\, dx = -e^{-x}\,dx.$
 October 21st, 2009, 10:33 PM #5 Senior Member   Joined: Feb 2008 Posts: 100 Thanks: 0 Re: Chain Rule QUestion I get it now, thanks.

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