Rules for manipulating derivatives

multivariable-calculus/manipulating-derivatives.md

@@ -1 +1,74 @@
-# Manipulating partial derivatives
+# Manipulating Partial Derivatives
+
+We sometimes need to find derivatives we don't obtain easily from one of these potentials. For this, we use some calculus rules.
+
+---
+
+#### Inversion
+
+If  x(y, z)  and  y(x, z), then:
+
+```{math}
+\left( \frac{\partial x}{\partial y} \right)_z = \frac{1}{\left( \frac{\partial y}{\partial x} \right)_z}
+```
+
+**Example:**
+
+```{math}
+Let:
+x = \frac{y^2}{z}
+```
+
+We can rearrange this to:
+```{math}
+y = \pm \sqrt{xz}
+```
+
+Now compute:
+```{math}
+\left( \frac{\partial x}{\partial y} \right)_z = \frac{2y}{z}
+```
+
+Then:
+```{math}
+\left( \frac{\partial y}{\partial x} \right)_z = \frac{z}{2y}
+```
+
+And confirm the inversion:
+```{math}
+\frac{1}{\left( \frac{\partial y}{\partial x} \right)_z} = \frac{2y}{z}
+```
+
+#### Chain Rule
+
+```{math}
+\left( \frac{\partial x}{\partial y} \right)_z = \left( \frac{\partial x}{\partial w} \right)_z \left( \frac{\partial w}{\partial y} \right)_z
+```
+**Example:**
+
+```{math}
+  w = yz  
+```
+ So,   
+ ```{math}
+ y = \frac{w}{z}
+ ```
+Then:
+```{math}
+x = \frac{(w/z)^2}{z} = \frac{w^2}{z^3}
+```
+
+Now compute:
+```{math}
+\left( \frac{\partial x}{\partial w} \right)_z = \frac{2w}{z^3}
+```
+
+And:
+```{math}
+\left( \frac{\partial w}{\partial y} \right)_z = z
+```
+
+Therefore:
+```{math}
+\left( \frac{\partial x}{\partial y} \right)_z = \frac{2w}{z^3} \cdot z = \frac{2w}{z^2} = \frac{2yz}{z^2} = \frac{2y}{z}
+```