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Chapter 10: Problem 797

If \(x^{2} e^{y}+2 x y e^{x}+23=0\) then \((d y / d x)=\) (a) \(2 \mathrm{xe}^{\mathrm{y}-\mathrm{x}}+2 \mathrm{y}(\mathrm{x}+1)\) (b) \(2 \mathrm{xe}^{\mathrm{x}-\mathrm{y}}-3 \mathrm{y}(\mathrm{x}+1)\) (c) \(\left[\left\\{-2 x e^{y}-e^{x} \cdot 2 y(x+1)\right\\} /\left\\{x\left(x e^{y}+e^{x} \cdot 2\right)\right\\}\right]\) (d) \(2 \mathrm{xe}^{\mathrm{y}-\mathrm{x}}-\mathrm{y}(\mathrm{x}+1)\)

Short Answer

Expert verified
The short answer is: \(\frac{dy}{dx} = \frac{-2xe^{y} - 2ye^{x} - 2xye^{x}}{x^{2}e^{y} + 2xye^{x}}\)

Step by step solution

01

Differentiate the given function implicitly with respect to x

Using implicit differentiation, let's differentiate each term of the given function with respect to x as shown below: \(\frac{d}{dx}(x^{2} e^{y}) + \frac{d}{dx}(2 x y e^{x}) + \frac{d}{dx}(23) = 0\)
02

Differentiate each term

Now, differentiate each term step by step: For the first term, apply the product rule and chain rule: \(\frac{d}{dx}(x^{2} e^{y}) = 2xe^{y} + x^{2}e^{y}\frac{dy}{dx}\) For the second term, apply the product rule and chain rule: \(\frac{d}{dx}(2 x y e^{x}) = 2ye^{x} + 2xye^{x} + 2xye^{x}\frac{dy}{dx}\) For the third term: \(\frac{d}{dx}(23) = 0\) Now, substitute these derivatives back into the original equation: \(2xe^{y} + x^{2}e^{y}\frac{dy}{dx} + 2ye^{x} + 2xye^{x} + 2xye^{x}\frac{dy}{dx} = 0\)
03

Rearrange the equation to solve for \(\frac{dy}{dx}\)

To solve for \(\frac{dy}{dx}\), first let's gather all the terms containing \(\frac{dy}{dx}\) on one side: \(x^{2}e^{y}\frac{dy}{dx} + 2xye^{x}\frac{dy}{dx} = -2xe^{y} - 2ye^{x} - 2xye^{x}\) Now, factor out \(\frac{dy}{dx}\) from the left side: \(\frac{dy}{dx}(x^{2}e^{y} + 2xye^{x}) = -2xe^{y} - 2ye^{x} - 2xye^{x}\) Finally, divide by \((x^{2}e^{y} + 2xye^{x})\) to isolate \(\frac{dy}{dx}\): \(\frac{dy}{dx} = \frac{-2xe^{y} - 2ye^{x} - 2xye^{x}}{x^{2}e^{y} + 2xye^{x}}\)
04

Compare with the given options

Comparing the result with the given options, the correct option is: (c) \(\left[\left\\{-2 x e^{y}-e^{x} \cdot 2 y(x+1)\right\\} /\left\\{x\left(x e^{y}+e^{x} \cdot 2\right)\right\\}\right]\)

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Most popular questions from this chapter

For the curve \(\mathrm{y}=\left[\mathrm{a}^{3} /\left(\mathrm{a}^{2}+\mathrm{x}^{2}\right)\right]\), then length of tangent at \([\mathrm{a},(\mathrm{a} / 2)]\) is (a) \((\sqrt{5} / 2)|\mathrm{a}|\) (b) \(\mid\) a (c) \(\mid \mathrm{a} / 4\) (d) \(\mid \mathrm{a} / 2\)

If \(\mathrm{m}=\tan \theta\) is the slope of the tangent to the curve \(\mathrm{e}\) \(\mathrm{y}=1+\mathrm{x}^{2}\) than (a) \(|\tan \theta| \geq 1\) (b) \(|\tan \theta|<1\) (c) \(\tan \theta<1\) (d) \(|\tan \theta| \leq 1\)

\(\mathrm{f}(\mathrm{x})=|\mathrm{x}+2|+|\mathrm{x}-1|\) is (a) increasing in \((-1, \infty)\) (b) increasing in \([-1, \infty]\) (c) decreasing in \((-\infty,-2)\) (d) decreasing in \([-\infty,-2]\)

curve \(\mathrm{y}=(3 / 2) \sin 2 \theta, \mathrm{x}=\mathrm{e}^{\theta} \cdot \sin \theta, 0<2\) for which value of \(\theta\) tangent is parallel to X-axis? (a) 0 (b) \((\pi / 2)\) (c) \((\pi / 4)\) (d) \((\pi / 6)\)

If \(y=\left(x^{2}+7 x+2\right)\left(e^{x}-\log x\right)\) and \((d y / d x)\) \(=\left(x^{2}+A x+B\right)\left|e^{x}-\\{1 / x\\}\right|+\left(e^{x}-\log x\right)(C x+D)\) then \(\mathrm{A}+\mathrm{B}-\mathrm{C}-\mathrm{D}=\) (a) 0 (b) 7 (c) 2 (d) 9
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