Question

Finding a derivative (sin(x)cos(y))² = 2.

a) y' = 1/(2cos(x)sin(y))
b) y' = cos(x)/sin(y)
c) y' = -sin(x)/cos(y)
d) y' = -cos(x)cos(y)/sin(y)

Answer 1

To find the derivative of the function (sin(x)cos(y))^2 = 2, implicit differentiation is applied. After simplifying, the derivative with respect to y, denoted as y', is -cos(x)/sin(y).

Step-by-step explanation:

The student is asking to find the derivative of the function (sin(x)cos(y))^2 = 2. This requires implicit differentiation because the function involves two variables, x and y, where y is a function of x (indicated by y'). Thus, we differentiate both sides with respect to x, applying the chain rule and the product rule where necessary.

Let's differentiate the left side with respect to x: 2(sin(x)cos(y))(cos(y)sin'(x) - sin(x)sin(y)y') = d/dx(2). The right side differentiates to 0 since the derivative of a constant is zero.

Now, we solve for y':
y' = -cos(x)sin(y) / (sin(x)cos^2(y))

To simplify further, we divide both the numerator and denominator by sin(y), which gives us:
y' = -cos(x) / cos^2(y).

This is equivalent to:
y' = -cos(x)/sin(y) because 1/cos(y) is sec(y), and sec(y) squared is 1/sin^2(y), which simplifies to -cos(x)/sin(y).