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Calculus Review of Trigonometric Identities

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Implicit Differentiation

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$\sin(xy^2) = 1 - x^2$

$\frac{dy}{dx} = \frac{-2x}{y^2\cos(xy^2)}$

$\cos(xy) = y^2 - x^2$

$\frac{dy}{dx} = \frac{2x - y \sin(xy)}{y(\sin(xy) + 2y)}$

$(x+y)^2 = 4$

$\frac{dy}{dx} = -\frac{2x+2y}{2x+2y}$

$x^3 + y^3 = 3xy$

$\frac{dy}{dx} = \frac{x^2 - y}{x - y^2}$

$\ln(y) + y = x + 5$

$\frac{dy}{dx} = \frac{1}{1 + 1/y}$

$ye^x = xe^y$

$\frac{dy}{dx} = \frac{ye^x - e^y}{xe^y - e^x}$

$xy + x^2 = 7$

$\frac{dy}{dx} = -\frac{2x+y}{x}$

$y \cos(x) = x^3 + 4$

$\frac{dy}{dx} = -\frac{x^2}{\cos(x)} - \frac{y \sin(x)}{\cos(x)}$

$y \ln(x) = x \ln(y)$

$\frac{dy}{dx} = \frac{y}{x} - \frac{\ln(y)}{\ln(x)}$

$y^5 - 5y = x^5$

$\frac{dy}{dx} = \frac{5x^4}{5y^4 - 5}$

$\sin(x) = \ln(yx)$

$\frac{dy}{dx} = \frac{y - \sin(x)}{yx}$

$x^2y^2 = 1$

$\frac{dy}{dx} = -\frac{x}{y}$

$x^3 + y^3 = 1$

$\frac{dy}{dx} = -\frac{x^2}{y^2}$

$e^y + y = x$

$\frac{dy}{dx} = \frac{1}{e^y+1}$

$\tan(y) = 3x$

$\frac{dy}{dx} = \frac{3}{\sec^2(y)}$

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