𝑧
=
𝑒
𝑦
+
2
𝑥
𝑦
+
𝑥
3
−
ln
⁡
𝑥
, find
∂
𝑦
∂
𝑥
when z = c given c is a constant.

Question

𝑧
=
𝑒
𝑦
+
2
𝑥
𝑦
+
𝑥
3
−
ln
⁡
𝑥
 , find 
∂
𝑦
∂
𝑥
 when z = c given c is a constant.

BlackTom AI · Accepted Answer

We begin by analyzing the given relation z = e^y + 2xy + x^3 − ln x with the constraint z = c being constant. This means that as x changes along the level curve z = c, the function z does not change, so dz/dx = 0 along that path.

Next, apply the chain rule for z as a function of x and y, where y is itself a function of x along the level curve. The total derivative with respect to x is:
dz/dx = (∂z/∂x) + (∂z/∂y)·(dy/dx).
Since dz/dx = 0, we have 0 = (∂z/∂x) + (∂z/∂y)·(dy/dx), which can be rearranged to solve for dy/dx:
dy/dx = − (∂z/∂x) / (∂z/∂y).

Compute the partial derivatives from z = e^y + 2xy + x^3 − ln x:
- ∂z/∂x = 2y + 3x^2 − (1/x).
- ∂z/∂y = e^y + 2x.

Substitute these into the formula for dy/dx:
dy/dx = − [2y + 3x^2 − (1/x)] / [e^y + 2x].

A few potential sources of confusion to watch for: (1) ensure the derivative of −ln x is −1/x, not −ln(x) or something similar; (2) remember that e^y differentiates with respect to y, and then you multiply by dy/dx when applying the chain rule; (3) the denominator cannot be zero on the level curve, otherwise dy/dx would be undefined at those points.

Note on the available answer options: in this instance, the provided set of answer choices is empty, so we cannot match a specific option to the derived expression. If a multiple-choice list were given, you would compare each option to the expression above, checking algebraic simplifications and ensuring the sign and terms align with dy/dx = −(2y + 3x^2 − 1/x)/(e^y + 2x).

In summary, the derivative of y with respect to x along the level curve z = c is dy/dx = −(2y + 3x^2 − 1/x) / (e^y + 2x).

𝑧 = 𝑒 𝑦 + 2 𝑥 𝑦 + 𝑥 3 − ln ⁡ 𝑥 , find ∂ 𝑦 ∂ 𝑥 when z = c given c is a constant.单项选择题

类似问题

In the equation below, 𝐶 is some constant. 17 𝑥 𝑒 𝑦 − 3 𝑦 = 𝐶 + 6 𝑥 Suppose the equation represents a curve passing through ( 2 , − 1 ) . Evaluate 𝑦 ′ at ( 2 , − 1 ) .

In the equation below, 𝐶 is some constant. 13 𝑥 𝑒 𝑦 − 3 𝑦 = 𝐶 + 4 𝑥 Suppose the equation represents a curve passing through ( 2 , − 1 ) . Evaluate 𝑦 ′ at ( 2 , − 1 ) .

At the given point, find the slope of the curve or the line that is tangent to the curve, as requested. 3x2y - π cos y = 4π, slope at (1, π)

Find dy/dx by implicit differentiation. x1/3 - y1/3 = 1

At the given point, find the line that is normal to the curve at the given point. 6x2y - π cos y = 7π, normal at (1, π)

Use implicit differentiation to find the slope of the tangent line to the curve defined by x y 4 + 6 x y = 5 at the point ( 5 , 1 ) .

Use implicit differentiation to find the slope of the tangent line to the curve defined by x y 2 + 2 x y = 5 at the point ( 3 , 1 ) .

更多留学生实用工具

智能学习助手

风格化写作助手

论文查重助手

文献引用助手

课堂转译助手

课堂笔记助手

Quiz搜索助手

学校历年真题

智能刷题助手

智能匹配练习

希望你的学习变得更简单