What is the slope of the tangent line of $(\frac{y}{x}) e^{\frac{x}{y} - x^{2}} = C$ $(y/x)e^(x/y-x^2)= C$ , where C is an arbitrary constant, at $(1, 2)$ $(1,2)$ ?

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What is the slope of the tangent line of $(\frac{y}{x}) e^{\frac{x}{y} - x^{2}} = C$ $(y/x)e^(x/y-x^2)= C$ , where C is an arbitrary constant, at $(1, 2)$ $(1,2)$ ?

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Answer 1 · 2018-06-02T01:35:14

10

Explanation:

Taking the logarithm of both sides gives us

$ln y - ln x + \frac{x}{y} - x^{2} = ln C$ $ln y -ln x +x/y-x^2 = ln C$

Differentiating with respect to $x$ $x$ yields

$\frac{1}{y} \frac{d y}{d x} - \frac{1}{x} + \frac{1}{y} - \frac{x}{y^{2}} \frac{d y}{d x} - 2 x = 0 \Rightarrow$ $1/y dy/dx -1/x +1/y-x/y^2dy/dx-2x = 0 implies$

$(\frac{1}{y} - \frac{x}{y^{2}}) \frac{d y}{d x} = 2 x + \frac{1}{x} - \frac{1}{y} \Rightarrow$ $(1/y-x/y^2)dy/dx =2x+1/x-1/y implies$

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

At the point $(1, 2)$ $(1,2)$ we have

$1 \times (2 - 1) \times \frac{d y}{d x} = (2 \times 1^{2} + 1) \times 2^{2} - 1 \times 2 \Rightarrow$ $1 times (2-1)times dy/dx = (2 times1^2+1)times 2^2-1times 2 implies$
$\frac{d y}{d x} = 10$ $dy/dx = 10$

Hence the slope of the tangent at $(1, 2)$ $(1,2)$ is 10

Note

I have assumed that tangent at $(1, 2)$ $(1,2)$ means tangent to the curve at $(1, 2)$ $(1,2)$ . Of course, this means that the constant $C$ $C$ can not be arbitrary - but rather has to be $\frac{2}{\sqrt{e}}$ $2/sqrt e$

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What is the slope of the tangent line of $(\frac{y}{x}) e^{\frac{x}{y} - x^{2}} = C$ $(y/x)e^(x/y-x^2)= C$ , where C is an arbitrary constant, at $(1, 2)$ $(1,2)$ ?

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What is the slope of the tangent line of (y/x)e^(x/y-x^2)= C (yx)exy−x2=C(y/x)e^(x/y-x^2)= C , where C is an arbitrary constant, at (1,2)(1,2)(1,2)?

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What is the slope of the tangent line of $(\frac{y}{x}) e^{\frac{x}{y} - x^{2}} = C$ $(y/x)e^(x/y-x^2)= C$ , where C is an arbitrary constant, at $(1, 2)$ $(1,2)$ ?