What is the implicit derivative of $1 = \frac{x}{y} - x tan y$ $1=x/y-xtany$ ?

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What is the implicit derivative of $1 = \frac{x}{y} - x tan y$ $1=x/y-xtany$ ?

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Answer 1 · 2015-12-13T20:44:09

$\frac{d y}{d x} = - \frac{y (y tan y - 1)}{x (1 + y^{2} {sec}^{2} y)}$ $dy/dx=-(y(ytany-1))/(x(1+y^2sec^2y))$

Explanation:

$\frac{d}{d x} [1 = \frac{x}{y} - x tan y]$ $d/dx[1=x/y-xtany]$

Find each derivative.

$\frac{d}{d x} [1] = 0$ $d/dx[1]=0$

$\frac{d}{d x} [\frac{x}{y}] = \frac{y \frac{d}{d x} [x] - x \frac{d}{d x} [y]}{y^{2}} = \frac{y - x \frac{d y}{d x}}{y^{2}}$ $d/dx[x/y]=(yd/dx[x]-xd/dx[y])/y^2=(y-xdy/dx)/y^2$

$\frac{d}{d x} [x tan y] = tan y \frac{d}{d x} [x] + x \frac{d}{d x} [tan y] = tan y + x {sec}^{2} y \frac{d y}{d x}$ $d/dx[xtany]=tanyd/dx[x]+xd/dx[tany]=tany+xsec^2ydy/dx$

Plug the derivatives back in.

$0 = \frac{y - x \frac{d y}{d x}}{y^{2}} - tan y - x {sec}^{2} y \frac{d y}{d x}$ $0=(y-xdy/dx)/y^2-tany-xsec^2ydy/dx$

$tan y = \frac{y - x \frac{d y}{d x}}{y^{2}} - x {sec}^{2} y \frac{d y}{d x}$ $tany=(y-xdy/dx)/y^2-xsec^2ydy/dx$

$y^{2} tan y = y - x \frac{d y}{d x} - x y^{2} {sec}^{2} y \frac{d y}{d x}$ $y^2tany=y-xdy/dx-xy^2sec^2ydy/dx$

$y^{2} tan y - y = \frac{d y}{d x} (- x - x y^{2} {sec}^{2} y)$ $y^2tany-y=dy/dx(-x-xy^2sec^2y)$

$\frac{d y}{d x} = - \frac{y (y tan y - 1)}{x (1 + y^{2} {sec}^{2} y)}$ $dy/dx=-(y(ytany-1))/(x(1+y^2sec^2y))$

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What is the implicit derivative of $1 = \frac{x}{y} - x tan y$ $1=x/y-xtany$ ?

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Explanation:

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What is the implicit derivative of $1 = \frac{x}{y} - x tan y$ $1=x/y-xtany$ ?