Partial differentiation

Partial derivatives help us analyse the influence of each individual variable on a system. This is crucial for understanding surfaces and optimisation problems in fields like engineering, physics and economics. Use this resource to learn how to find partial derivatives.

Partial derivatives reveal how a function with many variables changes when you adjust just one of the variables in the input. Consider the following equation:
\[z=x^{2}+4y^{2}\]

Here, we have two independent variables, so the function is \(f(x,y)\). The graph exists in a three-dimensional space.

When we differentiate such a function, we need to be able to differentiate \(z\) with respect to either \(x\) or \(z\). This involves treating the other variable as a constant. We calculate the partial derivatives:

• \(\dfrac{\partial z}{\partial x}\), the partial derivative of \(z\) with respect to \(x\) where we treat \(y\) as a constant
• \(\dfrac{\partial z}{\partial y}\), the partial derivative of \(z\) with respect to \(y\) where we treat \(x\) as a constant.

The symbol we use to denote a partial differential is the curve d, \(\partial\), often called "del".

An alternative notation for partial derivatives is \(f_{x}\) for \(\dfrac{\partial f}{\partial x}\)and \(f_{y}\) for \(\dfrac{\partial f}{\partial y}\).

Example 1 – evaluating partial derivatives

If \(z=x^{2}+4y^{2}\), find \(\dfrac{\partial z}{\partial x}\) and \(\dfrac{\partial z}{\partial y}\).

\(\dfrac{\partial z}{\partial x}\) is the partial derivative of \(z\) with respect to \(x\), treating \(y\) as a constant.
\[\begin{align*} \frac{\partial z}{\partial x} & = 2x+0\\
& = 2x
\end{align*}\]

\(\dfrac{\partial z}{\partial y}\) is the partial derivative of \(z\) with respect to \(y\), treating \(x\) as a constant.
\[\begin{align*} \frac{\partial z}{\partial y} & = 0+8y\\
& = 8y
\end{align*}\]

Evaluating partial derivatives at a point

We can find partial derivatives at particular points. For example, \(f_{x}(2,1)\) refers to the value of the partial derivative of \(f\) with respect to \(x\) at the point where \(x=2\) and \(y=1\).

Example – evaluating partial derivatives at a point

If \(f(x,y)=2x^{3}y+3y^{2}\), find \(f_{y}(1,3)\).

\(f_{y}\) indicates that we are looking for the derivative of \(f\) with respect to \(y\), treating \(x\) as a constant.
\[f_{y} = 2x^{3}+6y\]

We can then substitute \(x=1\) and \(y=3\) into \(f_{y}\) to find the partial derivative at that point.
\[\begin{align*} f_{y}(1,3) & = 2(1)^{3}+6(3)\\
& = 2+18\\
& = 20
\end{align*}\]

Higher order partial derivatives

A function \(z=f(x,y)\) has two partial derivatives: \(\dfrac{\partial f}{\partial x}\) and \(\dfrac{\partial f}{\partial y}\). This kind of function will have four second order partial derivatives.

1. Differentiating with respect to \(x\), then with respect to \(x\) gives us:
   \[\frac{\partial}{\partial x}\left(\frac{\partial f}{\partial x}\right)=\frac{\partial^{2}f}{\partial x^{2}}=f_{xx}\]
2. Differentiating with respect to \(y\), then with respect to \(y\) gives us:
   \[\frac{\partial}{\partial y}\left(\frac{\partial f}{\partial y}\right)=\frac{\partial^{2}f}{\partial y^{2}}=f_{yy}\]
3. Differentiating with respect to \(x\), then with respect to \(y\) gives us:
   \[\frac{\partial}{\partial y}\left(\frac{\partial f}{\partial x}\right)=\frac{\partial^{2}f}{\partial x\partial y}=f_{xy}\]
4. Differentiating with respect to \(y\), then with respect to \(x\) gives us:
   \[\frac{\partial}{\partial x}\left(\frac{\partial f}{\partial y}\right)=\frac{\partial^{2}f}{\partial y\partial x}=f_{yx}\]

\(f_{xy}\) is always equal to \(f_{yx}\).

Example – evaluating higher order partial derivatives

Find all second order partial derivatives of \(f(x,y)=x^{2}y^{3}+2x\cos(y)\).

First, we need to find the first order partial derivative with respect to \(x\) and \(y\). Remember, when we derive \(f\) with respect to \(x\), we treat all the \(y\) terms as constants. In this case, it would be \(y^{3}\) and \(2\cos(y)\).
\[f_{x}=2xy^{3}+2\cos(y)\]

For \(f_{y}\), we derive \(f\) with respect to \(y\), treating all the \(x\) terms as constants, i.e. \(x^{2}\) and \(2x\).
\[f_{y}=3x^{2}y^{2}-2x\sin(y)\]

The second order partial derivatives of \(f(x,y)\) are \(f_{xx}\), \(f_{yy}\) and \(f_{xy}=f_{yx}\).

For \(f_{xx}\), we find the derivative of \(f_{x}=2xy^{3}+2\cos(y)\) with respect to \(x\) and treat the \(y\) terms as a constant, i.e. \(y^{3}\) and \(2\cos(y)\).
\[f_{xx}=2y^{3}\]

For \(f_{xy}\), we find the derivative of \(f_{x}=2xy^{3}+2\cos(y)\) with respect to \(y\) and treat \(x\) as a constant, i.e. \(2x\).
\[f_{xy}=6xy^{2}-2\sin(y)\]

\(f_{xy}\) is the same as \(f_{yx}\), so \(f_{yx}=6xy^{2}-2\sin(y)\). We can prove this by deriving \(f_{y}=3x^{2}y^{2}-2x\sin(y)\) with respect to \(x\) while treating the \(y\) terms, \(3y^{2}\) and \(\sin(y)\), as constants.
\[f_{yx}=6xy^{2}-2\sin(y)\]

Finally, for \(f_{yy}\), we find the derivative of \(f_{y}=3x^{2}y^{2}-2x\sin(y)\) with respect to \(y\) while treating the \(x\) term, \(x^{2}\), as a constant.
\[f_{yy}=6x^{2}y-2x\cos(y)\]

Exercise – finding partial derivatives

1. Find the partial derivatives \(\dfrac{\partial z }{\partial x}\) and \(\dfrac{\partial z }{\partial y}\) for each of the following.
1. \(z=3x^{4}+2y^{3}\)
2. \(z=x^{2}y\)
3. \(z=3xe^{2y}\)
4. \(z=\ln(x^{3}y^{5}-2)\)
2. Find the value of the indicated partial derivative at the given point.
1. \(f_{x}(2,3)\) where \(f(x,y)=x^{4}-4y^{2}\)
2. \(f_{y}(-1,1)\) where \(f(x,y)=\ln(x^{2}+y^{3})\)
3. Find the first and second order partial derivatives of the following.
1. \(f(x,y)=x\ln(y)\)
2. \(f(x,y)=x^{3}+x^{2}y-3xy^{2}+y^{3}\)
3. \(f(x,y)=\sin(xy)\)
4. \(f(x,y)=x\cos(y)+ye^{x}\)

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