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Revision as of 23:12, 23 October 2008 edit152.1.222.41 (talk) →Gaaah, help← Previous edit Revision as of 23:24, 23 October 2008 edit undo152.1.222.41 (talk) →Find second derivativeNext edit →
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<math>\frac{d^{2}y}{dx^2} = \frac{-2-2\frac{x^2}{y^2}}{2y}\,</math><br /><br /> <math>\frac{d^{2}y}{dx^2} = \frac{-2-2\frac{x^2}{y^2}}{2y}\,</math><br /><br />
<math>\frac{d^{2}y}{dx^2} = -\frac{1}{y} - \frac{x^2}{y^3}\,</math><br /><br /> <math>\frac{d^{2}y}{dx^2} = -\frac{1}{y} - \frac{x^2}{y^3}\,</math><br /><br />
<math>\frac{d^{2}y}{dx^2} = -\frac{y^2}{y^3} - \frac{x^2}{y^3}\,</math><br /><br />
<math>\frac{d^{2}y}{dx^2} = -\frac{y^2+x^2}{y^3}\,</math><br /><br />
<math>\frac{d^{2}y}{dx^2} = -\frac{1}{y^3}\,</math><br /><br />

Revision as of 23:24, 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\,}

Gaaah, help~

Find d y d x {\displaystyle {\frac {dy}{dx}}\,} then find d 2 y d x 2 {\displaystyle {\frac {d^{2}y}{dx^{2}}}\,}

x 2 + y 2 = 1 {\displaystyle x^{2}+y^{2}=1\,}

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

Find first derivative

d y d x = − 2 x 2 y {\displaystyle {\frac {dy}{dx}}={\frac {-2x}{2y}}\,}

d y d x = − x y {\displaystyle {\frac {dy}{dx}}=-{\frac {x}{y}}\,}

Find second derivative

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

2 ( d y d x ) 2 + 2 y d 2 y d x 2 = − 2 {\displaystyle 2\left({\frac {dy}{dx}}\right)^{2}+2y{\frac {d^{2}y}{dx^{2}}}=-2\,}

2 ( − x y ) 2 + 2 y d 2 y d x 2 = − 2 {\displaystyle 2\left(-{\frac {x}{y}}\right)^{2}+2y{\frac {d^{2}y}{dx^{2}}}=-2\,}

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

2 y d 2 y d x 2 = − 2 − 2 x 2 y 2 {\displaystyle 2y{\frac {d^{2}y}{dx^{2}}}=-2-2{\frac {x^{2}}{y^{2}}}\,}

d 2 y d x 2 = − 2 − 2 x 2 y 2 2 y {\displaystyle {\frac {d^{2}y}{dx^{2}}}={\frac {-2-2{\frac {x^{2}}{y^{2}}}}{2y}}\,}

d 2 y d x 2 = − 1 y − x 2 y 3 {\displaystyle {\frac {d^{2}y}{dx^{2}}}=-{\frac {1}{y}}-{\frac {x^{2}}{y^{3}}}\,}

d 2 y d x 2 = − y 2 y 3 − x 2 y 3 {\displaystyle {\frac {d^{2}y}{dx^{2}}}=-{\frac {y^{2}}{y^{3}}}-{\frac {x^{2}}{y^{3}}}\,}

d 2 y d x 2 = − y 2 + x 2 y 3 {\displaystyle {\frac {d^{2}y}{dx^{2}}}=-{\frac {y^{2}+x^{2}}{y^{3}}}\,}

d 2 y d x 2 = − 1 y 3 {\displaystyle {\frac {d^{2}y}{dx^{2}}}=-{\frac {1}{y^{3}}}\,}