How do you find #dy/dx# by implicit differentiation given #y^2=ln(2x+3y)#?
Using the Chain Rule, we have,
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To find dy/dx by implicit differentiation for the equation y^2 = ln(2x + 3y), follow these steps:
 Differentiate both sides of the equation with respect to x.
 Use the chain rule when differentiating ln(2x + 3y) with respect to x.
 Solve the resulting equation for dy/dx.
Here's the solution:

Differentiating both sides: d/dx(y^2) = d/dx(ln(2x + 3y))

Applying the chain rule for the righthand side: 2y * (dy/dx) = (1/(2x + 3y)) * (d/dx(2x + 3y))

Simplify and solve for dy/dx: 2y * (dy/dx) = (1/(2x + 3y)) * (2 + 3(dy/dx))
Rearrange terms: 2y * (dy/dx) = (2 + 3(dy/dx)) / (2x + 3y)
Multiply both sides by (2x + 3y): 2y * (2x + 3y) * (dy/dx) = 2 + 3(dy/dx)
Expand and simplify: 4xy + 6y^2(dy/dx) = 2 + 3(dy/dx)
Bring all terms involving dy/dx to one side: 6y^2(dy/dx)  3(dy/dx) = 2  4xy
Factor out dy/dx: (6y^2  3) * (dy/dx) = 2  4xy
Divide both sides by (6y^2  3): dy/dx = (2  4xy) / (6y^2  3)
That's the derivative dy/dx for the given implicit equation y^2 = ln(2x + 3y).
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To find (\frac{dy}{dx}) by implicit differentiation given (y^2 = \ln(2x + 3y)), follow these steps:
 Differentiate both sides of the equation with respect to (x).
 Use the chain rule for differentiating (\ln(2x + 3y)).
 Solve for (\frac{dy}{dx}).
Differentiating both sides of (y^2 = \ln(2x + 3y)) with respect to (x):
[\frac{d}{dx}(y^2) = \frac{d}{dx}(\ln(2x + 3y))]
Applying the chain rule on the right side:
[2y\frac{dy}{dx} = \frac{1}{2x + 3y} \cdot (2 + 3\frac{dy}{dx})]
Simplify the equation:
[2y\frac{dy}{dx} = \frac{2 + 3\frac{dy}{dx}}{2x + 3y}]
Multiply both sides by (2x + 3y) to clear the fraction:
[2xy\frac{dy}{dx} = 2 + 3\frac{dy}{dx}]
Move all terms involving (\frac{dy}{dx}) to one side:
[2xy\frac{dy}{dx}  3\frac{dy}{dx} = 2]
Factor out (\frac{dy}{dx}):
[\frac{dy}{dx}(2xy  3) = 2]
Finally, solve for (\frac{dy}{dx}):
[\frac{dy}{dx} = \frac{2}{2xy  3}]
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When evaluating a onesided 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 onesided 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 onesided 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 onesided 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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