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

10

Taking the logarithm of both sides gives us

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To find the slope of the tangent line at the point ((1,2)) for the curve ( \frac{y}{x} e^{(x/y - x^2)} = C ), differentiate the equation implicitly with respect to (x) and (y).

Differentiating implicitly with respect to (x): [ \frac{y}{x} e^{(x/y - x^2)} + \frac{dy}{dx} e^{(x/y - x^2)} \left(1 - \frac{2x}{y}\right) = 0 ]

Differentiating implicitly with respect to (y): [ \frac{1}{x} e^{(x/y - x^2)} + \frac{y}{x^2} e^{(x/y - x^2)} \left(1 - \frac{x^2}{y^2}\right) \frac{dx}{dy} = 0 ]

Evaluate both equations at the point ((1,2)) to find (\frac{dy}{dx}) and (\frac{dx}{dy}).

Using these values, the slope (m) of the tangent line is given by: [ m = -\frac{dx}{dy} \times \frac{y}{\frac{dy}{dx}} ]

After calculating, the slope (m) at ((1,2)) is:

[ m = -\frac{dx}{dy} \times \frac{2}{\frac{dy}{dx}} ]

Solving for (m):

[ m = -\frac{e^{(1/2 - 1)}}{1 + \frac{4}{2^2}} \times \frac{2}{e^{(1/2 - 1)} \left(1 - \frac{2(1)}{2}\right)} ]

[ m = -\frac{e^{-1/2}}{1 + 1} \times \frac{2}{e^{-1/2} \left(1 - 1\right)} ]

[ m = -\frac{e^{-1/2}}{2} \times \frac{2}{0} ]

Since the denominator is 0, the slope is undefined at the point ((1,2)).

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To find the slope of the tangent line of the curve (y/x e^{x/y - x^2} = C) at the point (1,2), we first need to find the equation of the curve and then differentiate it implicitly with respect to (x).

Given: (y/x e^{x/y - x^2} = C)

Taking the natural logarithm of both sides, we get:

[\ln\left(\frac{y}{x} e^{x/y - x^2}\right) = \ln(C)]

Using logarithmic properties, we simplify to:

[\ln(y) - \ln(x) + \frac{x}{y} - x^2 = \ln(C)]

Now, differentiating implicitly with respect to (x), we get:

[\frac{1}{y} \frac{dy}{dx} - \frac{1}{x} - \frac{x}{y^2} \frac{dy}{dx} - 2x = 0]

At the point (1,2), we substitute (x = 1) and (y = 2) into the equation above to find the slope of the tangent line. After substituting, we solve for (\frac{dy}{dx}), which gives us the slope of the tangent line at that point.

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When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

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