# Find dz/dx and dz/dy if z=3x^2-2y^2+2xy .

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### 2 Answers

z = 3x^2-2y^2+2xy.

To find dz/dx we differentiate z with respect to x as a function of the variable x and we treat y as a constant.

dz/dx = d/dx (3x^2-2y^2+2xy).

dz/dx = d/dx(3x^2) - d/dx(2y^2) - d/dx(2xy)

dz/dx = 6x - 0 -2y.

Therefore dz/dx = 6x-2y.

To find dz/dy , we diffrentiate z withe respect to the variable y and treat x as constant.

Therefore dz/dy = d/dy {3x^2-2y^2 +2xy}

dz/dy = d/dy(3x^2) -d/dy(2y^2)- d/dy(2xy)

dz/dy = 0 - 2*2y -2x

dz/dy = -4y -2x.

We'll have to calculate partial derivative for given expression.

We'll calculate dz/dx.

z=3x^2-2y^2+2xy

We'll differentiate the expression of z with respect to x, treating y as a constant.

dz/dx = (d/dx)(3x^2-2y^2+2xy)

dz/dx = 3(d/dx)(x^2) - 2y^2(d/dx)(1) + 2y(d/dx)(x)

**dz/dx = 6x + 2y**

Now, we'll differentiate z with respect to y, treating x as a constant.

dz/dy = (d/dy)(3x^2-2y^2+2xy)

dz/dy = 3x^2(d/dy)(1) - 2(d/dy)(y^2) + 2x(d/dy)(y)

dz/dy = 0 - 4y + 2x

**dz/dy = 2x - 4y**