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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 /> | |||
| <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 /> | |||
| == Clock Problem ~ == | |||
| ===minute hand=== | |||
| <math>x=5\cos\left(\frac{\pi}{2}-\frac{t \cdot 2\pi}{60}\right)</math><br /><br /> | |||
| <math>y=5\sin\left(\frac{\pi}{2}-\frac{t \cdot 2\pi}{60}\right)</math><br /><br /> | |||
| ===hour hand=== | |||
| <math>x=4\cos\left(\frac{\pi}{2}-\frac{t \cdot 2\pi}{12}\right)</math><br /><br /> | |||
| <math>y=4\sin\left(\frac{\pi}{2}-\frac{t \cdot 2\pi}{12}\right)</math><br /><br /> | |||
| == Piston speed ~ == | |||
| <math>x_w = r\cos\left(\theta\right)</math><br /><br /> | |||
| <math>y_w = r\sin\left(\theta\right)</math><br /><br /> | |||
| <math>D_p = \sqrt{L^2-x_w^2}+y_w</math><br /><br /> | |||
| <math>\frac{dD_p}{dt} = \frac{1}{2}\left(L^2-x_w^2\right)^{-\frac{1}{2}}\cdot\left(-2x_w\frac{dx_w}{dt}\right)+\frac{dy_w}{dt}</math><br /><br /> | |||
| <math>\frac{dx_w}{dt} = -r\sin\left(\theta\right)\cdot\omega</math><br /><br /> | |||
| <math>\frac{dy_w}{dt} = r\cos\left(\theta\right)\cdot\omega</math><br /><br /> | |||
| <math>\theta = t\cdot\omega</math><br /><br /> | |||
| == Feon's Question 1 == | |||
| <math>y = \tan\left(\arcsin\left(x\right)\right)</math><br /><br /> | |||
| <math>\frac{dy}{dx} = \frac{\sec^2\left(\arcsin\left(x\right)\right)}{\sqrt{1-x^2}}</math><br /><br /> | |||
| ==Feon's Question 2== | |||
| <math>y = x\cdot\arcsec\left(x\right)</math><br /><br /> | |||
| <math>\frac{dy}{dx} = \arcsec\left(x\right) + \frac{x}{\left|x\right|\sqrt{x^2-1}}</math><br /><br /> | |||
| <math>\frac{dy}{dx} = \begin{cases} | |||
| \arcsec\left(x\right) + \frac{1}{\sqrt{x^2-1}}& \mbox{if }x>0 \\ | |||
| \arcsec\left(x\right) - \frac{1}{\sqrt{x^2-1}}& \mbox{if }x<0 | |||
| \end{cases}</math><br /><br /> | |||
| ==Feon's Question 3== | |||
| <math>y = x\left(\arcsin\left(x\right)\right)^2-2x+2\sqrt{1-x^2}</math><br /><br /> | |||
| <math>\frac{dy}{dx} = \left(\arcsin\left(x\right)\right)^2+x \cdot \frac{2\left(\arcsin\left(x\right)\right)}{\sqrt{1-x^2}}-2+2\cdot \frac{1}{2\sqrt{1-x^2}} \cdot -2x</math><br /><br /> | |||
| <math>\frac{dy}{dx} = \arcsin^2\left(x\right)+\frac{2x\arcsin\left(x\right)}{\sqrt{1-x^2}}-\frac{2x}{\sqrt{1-x^2}}-2</math><br /><br /> | |||