Understanding the Partial Derivative $\frac{\partial z}{\partial y}$

Consider the function f(x,y) = 1 + sin(x)cos(y)

Fix the value of

y

with this slider.y = -1.50

-5.00.0+5.0

Now let

x

vary along the curve of points with the fixed value of

y

above.x = 0.50

-5.00.0+5.0

1 + \sin(x) \cos(y)

(Drag to rotate)

Slice:

z = 1 + -1\sin(x)

If we let $y_0$ represent the fixed value for $y$ with the slider above, the height is now

1 + \cos(y_0)\sin(x) = 1 + -1 \sin(x)

and you can just take the derivative.

\frac{d}{dx}\left( 1 + -1\sin(x) \right) = -1\cos(x)

Conclusion: Computing $\frac{\partial z}{\partial x}$

This procedure works for any fixed value of $y$ . You could go up and change $y$ now and see all of the above computations update to that new fixed value. To compute the slope of that slice, all you need to do is treat $y$ as a constant.

\frac{\partial}{\partial x} \left( 1 + \sin(x)\cos(y) \right) = \cos(y) \cos(x).

Do you think you could go through the same steps to find $\frac{\partial z}{\partial y}$ ?

Understanding the Partial Derivative ∂z∂y\frac{\partial z}{\partial y}∂y∂z​

Conclusion: Computing ∂z∂x\frac{\partial z}{\partial x}∂x∂z​

Understanding the Partial Derivative $\frac{\partial z}{\partial y}$

Conclusion: Computing $\frac{\partial z}{\partial x}$