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Given that # sin(x/y) = 1/2 # find #dy/dx#?

Answer 1

Luke Alvarez

# dy/dx = y/x #

We have:

# sin(x/y) = 1/2 #

Differentiating Implicitly wrt #x# and applying the chain rule and the product rule we get:

# \ \ \ \ cos(x/y) {d/dx(x/y)} = 0 #

# :. cos(x/y) { (y)(d/dx x) - (d/dx y)(x) } / (y)^2 = 0 #

# :. cos(x/y) { (y)(1) - (dy/dx)(x) } / y^2 = 0 #

# :. cos(x/y) { y - x dy/dx } / y^2 = 0 #

# :. y - x dy/dx = 0 #

# :. x dy/dx = y #

# :. dy/dx = y/x #

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Answer 2

William Abdalla

To find ( \frac{dy}{dx} ) from the equation ( \sin\left(\frac{x}{y}\right) = \frac{1}{2} ), we use implicit differentiation:

Differentiating both sides with respect to ( x ), we get:

[ \frac{d}{dx} \left( \sin\left(\frac{x}{y}\right) \right) = \frac{d}{dx} \left( \frac{1}{2} \right) ]

Using the chain rule, we have:

[ \cos\left(\frac{x}{y}\right) \cdot \frac{1}{y} \cdot \frac{dy}{dx} = 0 ]

Solving for ( \frac{dy}{dx} ), we get:

[ \frac{dy}{dx} = 0 ]

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Answer from HIX Tutor

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.