In the equation below, C is some constant.

13 x e^{y} - 5 y = C + 6 x

Suppose the equation represents a curve passing through (2, -1). Evaluate y' at (2, -1).

Question

In the equation below, C is some constant.

13 x e^{y} - 5 y = C + 6 x

Suppose the equation represents a curve passing through (2, -1). Evaluate y' at (2, -1).

BlackTom AI · Accepted Answer

The problem gives an implicit relation: 13 x e^{y} - 5 y = C + 6 x, where C is a constant, and tells us the curve passes through (2, -1). To find y', we need to differentiate implicitly with respect to x.

First, differentiate both sides term by term. The left-hand side contains a product 13 x e^{y} and a linear term -5 y:
- The derivative of 13 x e^{y} with respect to x requires the product rule: d/dx [13 x e^{y}] = 13 e^{y} + 13 x e^{y} y'. Here e^{y} is a function of y, which itself depends on x, so we multiply by y' when differentiating e^{y}.
- The derivative of -5 y with respect to x is -5 y'.
The right-hand side derivative is simply 6, since C is a constant and d/dx (6x) = 6.

Putting these together gives:
13 e^{y} + 13 x e^{y} y' - 5 y' = 6.

Now collect the y' terms on one side and the constants on the other:
13 x e^{y} y' - 5 y' = 6 - 13 e^{y}.
Factor y' on the left:
y' (13 x e^{y} - 5) = 6 - 13 e^{y}.
Hence the general expression for y' is:
y' = (6 - 13 e^{y}) / (13 x e^{y} - 5).

Next, use the point (2, -1) to evaluate y'. At y = -1, we have e^{y} = e^{-1}.
Compute the numerator: 6 - 13 e^{y} = 6 - 13 e^{-1}.
Compute the denominator: 13 x e^{y} - 5 = 13 · 2 · e^{-1} - 5 = 26 e^{-1} - 5.
Therefore, y' at (2, -1) is:
y' = (6 - 13 e^{-1}) / (26 e^{-1} - 5).

Note that some equivalent algebraic forms are possible. For example, multiplying numerator and denominator by e converts the expression to (6e - 13) / (26 - 5e). The form shown above directly reflects the differentiation steps and substitution using the implicit relation.

In the absence of answer choices, this is the step-by-step route: differentiate with product rule for 13 x e^{y}, collect y' terms, solve for y', then substitute x = 2 and y = -1 by evaluating e^{y} = e^{-1}.

In the equation below, C is some constant. 13 x e^{y} - 5 y = C + 6 x Suppose the equation represents a curve passing through (2, -1). Evaluate y' at (2, -1).单项选择题

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