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Revision as of 22:11, 23 October 2008 editHayson1991 (talk | contribs)242 edits →9← Previous edit Revision as of 22:58, 23 October 2008 edit undo152.1.222.41 (talk) →Multiple u's: new sectionNext edit →
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<math>\frac{dy}{dx} = \frac{-2xy - y^2}{x^{2} + 2xy}</math><br /><br /> <math>\frac{dy}{dx} = \frac{-2xy - y^2}{x^{2} + 2xy}</math><br /><br />
<math>\frac{dy}{dx} = -\frac{2xy + y^2}{x^{2} + 2xy}</math><br /><br /> <math>\frac{dy}{dx} = -\frac{2xy + y^2}{x^{2} + 2xy}</math><br /><br />

== Multiple u's ==

To Find dy/dx for<br />
<math>y = 2\cos\left(\left(5x\right)^2\right)</math><br /><br />
===The way she explains it===
you'll make 3 u's<br />
<math>\text{Let }u = 2\cos\left(u\right)</math><br /><br />
<math>\text{Let }u = u^2\,</math><br /><br />
<math>\text{Let }u = 5x\,</math><br /><br />
===The way you should do it===
<math>\text{Let }y=2\cos\left(u\right)\,</math><br /><br />
<math>\text{Let }u=v^2\,</math><br /><br />
<math>\text{Let }v=5x\,</math><br /><br />
<math>\frac{dy}{dx}=\frac{dy}{du}*\frac{du}{dv}*\frac{dv}{dx}</math><br /><br />

Revision as of 22:58, 23 October 2008

x = tan ⁡ ( y ) {\displaystyle x=\tan \left(y\right)}

1 = sec 2 ⁡ ( y ) ∗ d y d x {\displaystyle 1=\sec ^{2}\left(y\right)*{\frac {dy}{dx}}} (Chain rule, derivative of tan=sec^2)

1 sec 2 ⁡ ( y ) = d y d x {\displaystyle {\frac {1}{\sec ^{2}\left(y\right)}}={\frac {dy}{dx}}}

cos 2 ⁡ ( y ) = d y d x {\displaystyle \cos ^{2}\left(y\right)={\frac {dy}{dx}}}

d y d x = cos 2 ⁡ ( y ) {\displaystyle {\frac {dy}{dx}}=\cos ^{2}\left(y\right)}

9~

x 2 y + x y 2 = 6 {\displaystyle x^{2}y+xy^{2}=6\,}

( 2 x ∗ y + x 2 ∗ d y d x ) + ( 1 ∗ y 2 + x ∗ 2 y d y d x ) = 0 {\displaystyle \left(2x*y+x^{2}*{\frac {dy}{dx}}\right)+\left(1*y^{2}+x*2y{\frac {dy}{dx}}\right)=0}

2 x y + x 2 d y d x + y 2 + 2 x y d y d x = 0 {\displaystyle 2xy+x^{2}{\frac {dy}{dx}}+y^{2}+2xy{\frac {dy}{dx}}=0}

x 2 d y d x + 2 x y d y d x = − 2 x y − y 2 {\displaystyle x^{2}{\frac {dy}{dx}}+2xy{\frac {dy}{dx}}=-2xy-y^{2}}

d y d x = − 2 x y − y 2 x 2 + 2 x y {\displaystyle {\frac {dy}{dx}}={\frac {-2xy-y^{2}}{x^{2}+2xy}}}

d y d x = − 2 x y + y 2 x 2 + 2 x y {\displaystyle {\frac {dy}{dx}}=-{\frac {2xy+y^{2}}{x^{2}+2xy}}}

Multiple u's

To Find dy/dx for
y = 2 cos ⁡ ( ( 5 x ) 2 ) {\displaystyle y=2\cos \left(\left(5x\right)^{2}\right)}

The way she explains it

you'll make 3 u's
Let  u = 2 cos ⁡ ( u ) {\displaystyle {\text{Let }}u=2\cos \left(u\right)}

Let  u = u 2 {\displaystyle {\text{Let }}u=u^{2}\,}

Let  u = 5 x {\displaystyle {\text{Let }}u=5x\,}

The way you should do it

Let  y = 2 cos ⁡ ( u ) {\displaystyle {\text{Let }}y=2\cos \left(u\right)\,}

Let  u = v 2 {\displaystyle {\text{Let }}u=v^{2}\,}

Let  v = 5 x {\displaystyle {\text{Let }}v=5x\,}

d y d x = d y d u ∗ d u d v ∗ d v d x {\displaystyle {\frac {dy}{dx}}={\frac {dy}{du}}*{\frac {du}{dv}}*{\frac {dv}{dx}}}