My Math Forum Chain rule questoin

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 January 3rd, 2009, 10:26 AM #1 Newbie   Joined: Jan 2009 Posts: 1 Thanks: 0 Chain rule questoin Hello all, I'm stuck on a problem and I'm hoping someone can help. Here is the problem: If y = f((x^2 + 9)^1/2) and f '(5) = -2, find dy/dx when x = 4. Here is what I have tried so far. I know that dy/dx = dy/du * du/dx. If u = (x^2 + 9)^1/2, then du/dx = x/(x^2 + 9)^1/2. so... -2 = dy/du * x/(x^2 + 9)^1/2 If I substitute x = 5, then... -2 = dy/du * 5/(34)^1/2 This doesn't seem right. I think I'm going down the wrong path to solve this one. Can anyone help me with this problem? Thanks.
 January 3rd, 2009, 06:04 PM #2 Senior Member   Joined: Dec 2008 Posts: 251 Thanks: 0 Re: Chain rule questoin Remember that when $x\,=\,5$, $f'(\sqrt{x^2\,+\,9})\,=\,f#39;(\sqrt{34})$, which is not necessarily equal to $f'(5)$.
 January 4th, 2009, 11:22 AM #3 Member   Joined: Dec 2008 From: Washington state, of the United States Posts: 37 Thanks: 0 Re: Chain rule questoin $y= f(\sqrt{x^2 + 9})$ and $f '(5) = -2$, find dy/dx when x = 4 $y= f(\sqrt{x^2 + 9})$ $y'= f#39;(\sqrt{x^2 + 9}) \cdot \frac{d}{dx} \sqrt{x^2 + 9}$ $y'= f#39;(\sqrt{x^2 + 9}) \cdot \frac{1}{2\cdot \sqrt{x^2 + 9}} \cdot \frac{d}{dx} (x^2 + 9)$ $y'= f#39;(\sqrt{x^2 + 9}) \cdot \frac{1}{2\cdot \sqrt{x^2 + 9}} \cdot 2x$ Let x = 4 $y'= f#39;(\sqrt{4^2 + 9}) \cdot \frac{1}{2\cdot \sqrt{4^2 + 9}} \cdot 2(4)$ $y'= f#39;(\sqrt{25}) \cdot \frac{8}{2\cdot \sqrt{25}}$ $y'= f#39;(5) \cdot \frac{4}{5}$ since f'(5) = -2: $y' = -2 \cdot \frac{4}{5}$ $y' = -\frac{8}{5}$

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