Question

How to find the first partial derivatives ?:
\[f\left( x,y,z,t \right)=\dfrac{xy}{t+2z}\]

Answer 1

Hint: The first and the most important rule that we should keep in mind while finding out the partial derivative is that when we find the partial derivative of the function with respect to one variable, then we should treat other variable/ variables as a constant. For example, if we have to find the partial derivative of a function of \[x\] and $y$ with respect to \[x\], then $y$ is treated as a constant. And then the function is derived in the same way as normal differentiation.

Complete step by step solution:
We are given the function \[f\left( x,y,z,t \right)=\dfrac{xy}{t+2z}\]
First of all, let us find the partial derivative of $f$ with respect to \[x\]
For this, we need to treat $y$, $z$, and $t$ as constant and then differentiate the function normally with respect to \[x\].
$\dfrac{\text{ }\partial f}{\partial x}=\dfrac{\partial (\dfrac{xy}{t+2z})}{\partial x}$
Since $y$, $z$, and $t$ are treated as constant so it can be taken out of the derivative without any changes.
This implies , $\dfrac{\text{ }\partial f}{\partial x}=\dfrac{y}{t+2z}\dfrac{\partial (x)}{\partial x}$
Since $\dfrac{\partial x}{\partial x}$ gives $1$
Therefore ${{f}_{x}}=\dfrac{\text{ }\partial f}{\partial x}=y/t+2z$
Now, let us find the partial derivative of $f$ with respect to $y$
For this, we need to treat \[x\], $z$ and $t$ as constant and then differentiate the function normally with respect to $y$.
$\dfrac{\text{ }\partial f}{\partial y}=\dfrac{\partial (\dfrac{xy}{t+2z})}{\partial y}$
Since \[x\], $z$ and $t$ are treated as constants so it can be taken out of the derivative without any changes.
This implies , $\dfrac{\text{ }\partial f}{\partial y}=\dfrac{x}{t+2z}\dfrac{\partial (y)}{\partial y}$
Since $\dfrac{\partial y}{\partial y}$ gives $1$
Therefore ${{f}_{y}}=\dfrac{\text{ }\partial f}{\partial y}=x/t+2z$
Now, let us find the partial derivative of $f$ with respect to $z$
For this, we need to treat \[x\], $y$, and $t$ as constant and then differentiate the function normally with respect to $z$.
$\dfrac{\text{ }\partial f}{\partial z}=\dfrac{\partial (\dfrac{xy}{t+2z})}{\partial z}$
Since \[x\], $y$ and $t$ are treated as constants so it can be taken out of the derivative without any changes.
This implies , $\dfrac{\text{ }\partial f}{\partial z}=xy\dfrac{\partial {{(t+2z)}^{-1}}}{\partial z}$
Since by chain rule , $\dfrac{\partial {{x}^{n}}}{\partial x}=n{{(x)}^{(n-1)}}$
Thus $\dfrac{\partial ({{(t+2z)}^{-1}}}{\partial z}=(-1){{(t+2z)}^{-2}}(2)$
Therefore ${{f}_{z}}=\dfrac{\text{ }\partial f}{\partial z}=-2xy/{{(t+2z)}^{2}}$
Now, let us find the partial derivative of $f$ with respect to $t$
For this, we need to treat \[x\], $y$, and $z$ as constant and then differentiate the function normally with respect to $t$.
$\dfrac{\text{ }\partial f}{\partial t}=\dfrac{\partial (\dfrac{xy}{t+2z})}{\partial t}$
Since \[x\], $y$ and $z$ are treated as constants so it can be taken out of the derivative without any changes.
This implies , $\dfrac{\text{ }\partial f}{\partial t}=xy\dfrac{\partial {{(t+2z)}^{-1}}}{\partial t}$
Since by chain rule , $\dfrac{\partial {{x}^{n}}}{\partial x}=n{{(x)}^{(n-1)}}$
Thus $\dfrac{\partial ({{(t+2z)}^{-1}}}{\partial t}=(-1){{(t+2z)}^{-2}}$

Therefore ${{f}_{t}}=\dfrac{\text{ }\partial f}{\partial t}=-xy/{{(t+2z)}^{2}}$

Note:
To make this more simple , we rewrite the equation with capital letters except in the variable we're differentiating with respect to, can help. For example
\[{{f}_{z}}\] is more easily calculated if you write
\[f=XY/T+2z\]. Definitely, now it is more clear that \[x\], $y$, t are constants.