# How to Find the first partial derivatives?:

## $f \left(x , y , z , t\right) = \frac{x y}{t + 2 z}$

May 1, 2018

${f}_{x} = \frac{y}{t + 2 z}$
${f}_{y} = \frac{x}{t + 2 z}$
${f}_{z} = \frac{- 2 x y}{t + 2 z} ^ 2$
${f}_{t} = \frac{- x y}{t + 2 z} ^ 2$

#### Explanation:

When you find the partial derivative of a function with respect to a particular variable (say, $x$), then you treat every other variable like it's a constant. You then perform the derivative exactly as before. It takes a few problems before it "clicks", but the process will soon feel exactly the same as regular differentiation.

We have $f \left(x , y , z , t\right) = \frac{x y}{t + 2 z}$.

We'll first find $\frac{\partial f}{\partial x}$, which can be more conveniently notated ${f}_{x}$. Both notations refer to the first partial derivative of $f$ with respect to $x$. For ${f}_{x}$, we treat $x$ like a variable and everything else like a regular number. Thus, $f = \left(\frac{y}{t + 2 z}\right) \left(x\right)$ and the leftmost term is considered constant. Because the derivative of the function $C x$ is $C$, where $C$ is constant, it follows that ${f}_{x} = \frac{y}{t + 2 z}$.

A similar procedure is followed to find ${f}_{y}$. Since $f = \left(\frac{x}{t + 2 z}\right) \left(y\right)$, it follows that ${f}_{y} = \frac{x}{t + 2 z}$.

${f}_{z}$ is found by seeing that $f = \left(x y\right) {\left(t + 2 z\right)}^{- 1}$. (Remember, $x , y , t$ are constant.) Thus ${f}_{z} = \left(- 1\right) \left(x y\right) {\left(t + 2 z\right)}^{- 2} \left(2\right)$. Note that we used the chain rule. This can be simplified as ${f}_{z} = \frac{- 2 x y}{t + 2 z} ^ 2$.

Just as we found ${f}_{z}$, ${f}_{t} = \left(- 1\right) \left(x y\right) {\left(t + 2 z\right)}^{- 2}$. This can be written ${f}_{t} = \frac{- x y}{t + 2 z} ^ 2$.

If you still find the process difficult, simply rewriting the equation with capital letters except in the variable you're differentiating with respect to, can help. For example ${f}_{z}$ is more easily calculated if you write $f = \frac{X Y}{T + 2 z}$. Mentally, it is more clear that $x , y , t$ are constants.