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Find the derivative of y'(x)
(I) xe^y- 3ysinx =1

asked
User Lual
by
8.0k points

1 Answer

3 votes

Explanation:

To find the derivative of the given equation with respect to x, we'll use implicit differentiation. Let's go step by step:

Given equation: xe^y - 3y*sin(x) = 1

1. Differentiate both sides with respect to x:

Differentiate the left-hand side of the equation using the product rule and chain rule:

d/dx (xe^y) - d/dx (3y*sin(x)) = d/dx(1)

e^y * (x * dy/dx + 1) - 3y' * sin(x) - 3y * cos(x) = 0

2. Solve for y':

Rearrange the equation to solve for y':

e^y * (x * dy/dx + 1) - 3y' * sin(x) - 3y * cos(x) = 0

e^y * x * dy/dx + e^y - 3y' * sin(x) - 3y * cos(x) = 0

Now isolate the term with y':

e^y * x * dy/dx = 3y' * sin(x) + 3y * cos(x) - e^y

Divide both sides by (e^y * x):

dy/dx = (3y' * sin(x) + 3y * cos(x) - e^y) / (e^y * x)

So, the derivative of y with respect to x, y'(x), is:

y'(x) = (3y' * sin(x) + 3y * cos(x) - e^y) / (e^y * x)

answered
User Pdross
by
7.5k points
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