How do you implicitly differentiate $2 = e^{x - y} cos y$ $2=e^(x-y)cosy$ ?

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How do you implicitly differentiate $2 = e^{x - y} cos y$ $2=e^(x-y)cosy$ ?

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Answer 1 · 2018-03-19T20:40:25

$\frac{d y}{d x} = \frac{cos y}{sin y + cos y}$ $dy/dx=(cosy)/(siny+cosy)$

Explanation:

$differentiate e^{x - y} cos y using the product rule$ $"differentiate "e^(x-y)cosy" using the "color(blue)"product rule"$

$0 = e^{x - y} (- sin y) \frac{d y}{d x} + cos y e^{x - y} (1 - \frac{d y}{d x})$ $0=e^(x-y)(-siny)dy/dx+cosye^(x-y)(1-dy/dx)$

$0 = - e^{x - y} sin y \frac{d y}{d x} + cos y e^{x - y} - cos y e^{x - y} \frac{d y}{d x}$ $color(white)(0)=-e^(x-y)sinydy/dx+cosye^(x-y)-cosye^(x-y)dy/dx$

$\Rightarrow \frac{d y}{d x} (- e^{x - y} sin y - cos y e^{x - y}) = - cos y e^{x - y}$ $rArrdy/dx(-e^(x-y)siny-cosye^(x-y))=-cosye^(x-y)$

$\Rightarrow \frac{d y}{d x} = \frac{- cos y e^{x - y}}{- e^{x - y} (sin y + cos y)} = \frac{cos y}{sin y + cos y}$ $rArrdy/dx=(-cosye^(x-y))/(-e^(x-y)(siny+cosy))=(cosy)/(siny+cosy)$

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How do you implicitly differentiate $2 = e^{x - y} cos y$ $2=e^(x-y)cosy$ ?

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How do you implicitly differentiate $2 = e^{x - y} cos y$ $2=e^(x-y)cosy$ ?