R revinnova New member Joined Sep 29, 2006 Messages 6 Oct 19, 2006 #1 I have a little trouble understanding the circled problem. The part I'm stuck on is highlighted in green. How did the study guide get that? [/url][/list]
I have a little trouble understanding the circled problem. The part I'm stuck on is highlighted in green. How did the study guide get that? [/url][/list]
S soroban Elite Member Joined Jan 28, 2005 Messages 5,584 Oct 19, 2006 #2 Hello, revinnova! How did the study guide get that? Click to expand... My queston: how come you didn't get that? (It's part of the Quotient Rule.) Differentiate: \(\displaystyle \L\, y\;=\;\frac{x}{a^2\sqrt{a^2\,-\,x^2}}\,+\,C\) Click to expand... We have: \(\displaystyle \L\:y \;= \;\frac{1}{a^2}\,\cdot\,\frac{x}{(a^2\,-\,x^2)^{1/2}}\,+\,C\) Then: \(\displaystyle \L\,\frac{dy}{dx}\;=\;\frac{1}{a^2}\,\cdot\,\frac{(a^2\,-\,x^2)^{\frac{1}{2}}\,\cdot\,1 \:-\:x\,\cdot\,\left[\overbrace{\frac{1}{2}(a^2\,-\,x^2)^{-\frac{1}{2}}(-2x)}\right] }{a^2\,-\,x^2}\) Simplify the expression in brackets: . . \(\displaystyle \L\frac{1}{2}(a^2\,-\,x^2)^{-\frac{1}{2}}(-2x) \;=\;\frac{1}{\not{2}}\,\cdot\,\frac{1}{(a^2\,-\,x^2)^{\frac{1}{2}}}\,\cdot(-\not{2}x) \;=\;\frac{-x}{\sqrt{a^2\,-\,x^2}}\) Then we have: \(\displaystyle \L\:\frac{dy}{dx}\;=\;\frac{1}{a^2}\,\cdot\,\frac{\sqrt{a^2\,-\,x^2}\,-\,x\left(\frac{-x}{\sqrt{a^2\,-\,x^2}}\right) }{a^2\,-\,x^2}\) . . . Got it?
Hello, revinnova! How did the study guide get that? Click to expand... My queston: how come you didn't get that? (It's part of the Quotient Rule.) Differentiate: \(\displaystyle \L\, y\;=\;\frac{x}{a^2\sqrt{a^2\,-\,x^2}}\,+\,C\) Click to expand... We have: \(\displaystyle \L\:y \;= \;\frac{1}{a^2}\,\cdot\,\frac{x}{(a^2\,-\,x^2)^{1/2}}\,+\,C\) Then: \(\displaystyle \L\,\frac{dy}{dx}\;=\;\frac{1}{a^2}\,\cdot\,\frac{(a^2\,-\,x^2)^{\frac{1}{2}}\,\cdot\,1 \:-\:x\,\cdot\,\left[\overbrace{\frac{1}{2}(a^2\,-\,x^2)^{-\frac{1}{2}}(-2x)}\right] }{a^2\,-\,x^2}\) Simplify the expression in brackets: . . \(\displaystyle \L\frac{1}{2}(a^2\,-\,x^2)^{-\frac{1}{2}}(-2x) \;=\;\frac{1}{\not{2}}\,\cdot\,\frac{1}{(a^2\,-\,x^2)^{\frac{1}{2}}}\,\cdot(-\not{2}x) \;=\;\frac{-x}{\sqrt{a^2\,-\,x^2}}\) Then we have: \(\displaystyle \L\:\frac{dy}{dx}\;=\;\frac{1}{a^2}\,\cdot\,\frac{\sqrt{a^2\,-\,x^2}\,-\,x\left(\frac{-x}{\sqrt{a^2\,-\,x^2}}\right) }{a^2\,-\,x^2}\) . . . Got it?
R revinnova New member Joined Sep 29, 2006 Messages 6 Oct 20, 2006 #3 Yeah, I guess I was too lazy to calculate it. Yeah, I got it. Thanks!
