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Revision as of 23:02, 23 October 2008 edit74.183.242.181 (talk)No edit summary← Previous edit Revision as of 23:02, 23 October 2008 edit undo74.183.242.181 (talk) →The way you should do itNext edit →
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<math>\text{Let }u = u^2\,</math><br /><br /> <math>\text{Let }u = u^2\,</math><br /><br />
<math>\text{Let }u = 5x\,</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 />


Find <math>\frac{dy}{dx}\,</math> then find <math>\frac{d^2y}{dx^2}\,</math> <br /><br /> Find <math>\frac{dy}{dx}\,</math> then find <math>\frac{d^2y}{dx^2}\,</math> <br /><br />

Revision as of 23:02, 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\,}


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\,}