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Revision as of 22:11, 23 October 2008 editHayson1991 (talk | contribs)242 edits →9← Previous edit Revision as of 22:11, 23 October 2008 edit undoHayson1991 (talk | contribs)242 edits →9Next edit →
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<math>\frac{dy}{dx} = \cos^2\left(y\right)</math><br /><br /> <math>\frac{dy}{dx} = \cos^2\left(y\right)</math><br /><br />


== 9 == == 9~ ==
<math>x^{2}y + xy^2 = 6\,</math><br /><br /> <math>x^{2}y + xy^2 = 6\,</math><br /><br />
<math>\left(2x*y + x^{2}*\frac{dy}{dx}\right) + \left(1*y^2 + x*2y\frac{dy}{dx}\right) = 0</math><br /><br /> <math>\left(2x*y + x^{2}*\frac{dy}{dx}\right) + \left(1*y^2 + x*2y\frac{dy}{dx}\right) = 0</math><br /><br />

Revision as of 22:11, 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}}}