| Revision as of 23:02, 23 October 2008 edit74.183.242.181 (talk) →The way you should do it← Previous edit | Latest revision as of 11:03, 6 July 2012 edit undoHayson1991 (talk | contribs)242 edits ←Blanked the page | ||
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| <math>x = \tan\left(y\right)</math><br /><br /> | |||
| <math>1 = \sec^2\left(y\right)*\frac{dy}{dx}</math> (Chain rule, derivative of tan=sec^2)<br /><br /> | |||
| <math>\frac{1}{\sec^2\left(y\right)} = \frac{dy}{dx}</math><br /><br /> | |||
| <math>\cos^2\left(y\right) = \frac{dy}{dx}</math><br /><br /> | |||
| <math>\frac{dy}{dx} = \cos^2\left(y\right)</math><br /><br /> | |||
| == 9~ == | |||
| <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>2xy + x^{2}\frac{dy}{dx} + y^2 + 2xy\frac{dy}{dx} = 0</math><br /><br /> | |||
| <math>x^{2}\frac{dy}{dx} + 2xy\frac{dy}{dx} = -2xy - y^2</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 /> | |||
| Find <math>\frac{dy}{dx}\,</math> then find <math>\frac{d^2y}{dx^2}\,</math> <br /><br /> | |||
| <math>x^{2} + y^{2} = 1\,</math><br /><br /> | |||