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Chapter 6: Problem 80

If \(e^{x+y}-x=0\), prove that \(\frac{d y}{d x}=\frac{1-x}{x}\).

Short Answer

Expert verified
Hence, it is proved that for the given equation \(e^{x+y}-x=0\), the derivative of \(y\) with respect to \(x\) is \(\frac{1-x}{x}\).

Step by step solution

01

Differentiate the equation

Start by differentiating the equation \(f(x,y) = e^{x+y}-x = 0\). Applying the chain rule on \(e^{x+y}\), \(\frac{df}{dx} = \frac{d}{dx} e^{x+y} - 1 = e^{x+y}(\frac{dx}{dx} + \frac{dy}{dx}) -1 = e^{x+y}(1 + \frac{dy}{dx}) - 1\). Here, \( \frac{dy}{dx}\) is the derivative of \(y\) with respect to \(x\). We want to find this.
02

Rearrange the equation to isolate the derivative

Setting the derivative of the function \(\frac{df}{dx} = 0\) since \(f(x,y) = 0\), we have \(0 = e^{x+y}(1 + \frac{dy}{dx}) - 1\). Rearranging terms to isolate \(\frac{dy}{dx}\), we get: \(\frac{dy}{dx}=\frac{1-e^{x+y}}{e^{x+y}}\).
03

Substitute the original equation

Since the original equation was \(e^{x+y}=x\), let's plug it into the equation from step 2 to get: \(\frac{dy}{dx} = \frac{1-x}{x}\).

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