Question

Найти dz/dx и dz/dy для y=cos^2x/2y

Answer 1

$z=cos^2\frac{x}{2y}\\\\z'_{x}=2\, cos\frac{x}{2y}\cdot sin\frac{x}{2y}\cdot \frac{1}{2y}=\frac{1}{2y}\cdot sin\frac{x}{y} \\\\z'_{y}=2\, cos\frac{x}{2y}\cdot sin\frac{x}{2y}\cdot \frac{x}{2}\cdot (-\frac{1}{y^2}) =-\frac{x}{2y^2}\cdot sin\frac{x}{y}$

Answer 2

To find dz/dx and dz/dy for the given expression y = cos^2(x)/2y, we'll first rewrite the equation in a more suitable form. Then, we can find the derivatives with respect to x and y separately.

Step 1: Rewrite the equation to isolate z: Start with the given equation: y = cos^2(x)/(2y)

Multiply both sides by 2y to get rid of the fraction on the right side: 2y^2 = cos^2(x)

Now, we can express z in terms of x and y: z = 2y^2 - cos^2(x)

Step 2: Find dz/dx and dz/dy:

Now, we'll differentiate z with respect to x and y.

a) dz/dx: To find dz/dx, we differentiate z with respect to x, treating y as a constant: dz/dx = d/dx (2y^2 - cos^2(x))

The derivative of 2y^2 with respect to x is 0, as it's a constant with respect to x.

Now, differentiate cos^2(x) with respect to x using the chain rule: d/dx (cos^2(x)) = -2cos(x)sin(x)

So, dz/dx = 0 - (-2cos(x)sin(x)) = 2cos(x)sin(x).

b) dz/dy: To find dz/dy, we differentiate z with respect to y, treating x as a constant: dz/dy = d/dy (2y^2 - cos^2(x))

The derivative of 2y^2 with respect to y is 4y.

The derivative of cos^2(x) with respect to y is 0 since cos^2(x) doesn't depend on y.

So, dz/dy = 4y - 0 = 4y.

Therefore, the derivatives are: dz/dx = 2cos(x)sin(x) dz/dy = 4y