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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 /> | |||
| == Gaaah, help == | |||
| 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 /> | |||
| <math>2x + 2y\frac{dy}{dx} = 0\,</math><br /><br /> | |||
| ===Find first derivative=== | |||
| <math>\frac{dy}{dx} = \frac{-2x}{2y}\,</math><br /><br /> | |||
| <math>\frac{dy}{dx} = -\frac{x}{y}\,</math><br /><br /> | |||
| ===Find second derivative=== | |||
| <math>2 + \left(2\frac{dy}{dx}*\frac{dy}{dx} + 2y*\frac{d^{2}y}{dx^2}\right) = 0\,</math><br /><br /> | |||
| <math>2\left(\frac{dy}{dx}\right)^2 + 2y\frac{d^{2}y}{dx^2} = -2\,</math><br /><br /> | |||
| <math>2\left(-\frac{x}{y}\right)^2 + 2y\frac{d^{2}y}{dx^2} = -2\,</math><br /><br /> | |||
| <math>2\frac{x^2}{y^2} + 2y\frac{d^{2}y}{dx^2} = -2\,</math><br /><br /> | |||
| <math>2y\frac{d^{2}y}{dx^2} = -2-2\frac{x^2}{y^2}\,</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 /> | |||