Partial derivatives follow directly from derivatives you have seen in single variable calculus. Calculation is pretty straightforward but, as is common in multivariable calculus, you need to watch your notation carefully.
To calculate partial derivatives, you are given a function, usually of more than one variable, and you are asked to take the derivative with respect to one of the variables. To do so, you consider the other variable as a constant. Before we go any further, let's discuss notation.
Remember from single variable calculus you are given a function, say \(f(x)\), and you are asked to take the derivative (with respect to x, of course, since x is the only variable ). For partial derivatives, you have more than one variable, say \(g(x,y)\). Here is a comparison of how you write the derivative of g with respect to x in both cases.
single variable  multivariable  

\(\displaystyle{ \frac{df}{dx} }\) or \(f'(x)\)  \(\displaystyle{ \frac{\partial g}{ \partial x} }\) or \(g_x(x,y)\) 
The \(d\) to indicate derivative for a single variable function is replaced by \( \partial \) for a partial derivative. This can also be written with a subscript to indicate the variable that we are taking the derivative with respect to. Notice that with partial derivatives, there is no 'prime' notation, since there is no way to determine the derivative variable. We need to show the variable somewhere to make any sense out of a partial derivative.
Example   Okay, we are ready for an example. Let's find both partial derivatives of \(g(x,y) = x^2y\), meaning \(\partial g/\partial x\) and \(\partial g/\partial y\).
partial derivative of \(g\) with respect to \(x\) \(\left[ \partial g/\partial x \right]\) 

\( y \) is a constant 
\(\displaystyle{ \frac{\partial g}{\partial x} = }\) \(\displaystyle{ \frac{\partial}{\partial x}[x^2y] = }\) \(\displaystyle{ y \frac{\partial}{\partial x}[x^2] = }\) \(\displaystyle{ y(2x) = 2xy }\) 
\(\displaystyle{\frac{\partial}{\partial x}[x^2y] = 2xy}\) 
partial derivative of \(g\) with respect to \(y\) \(\left[ \partial g/\partial y \right]\) 

\( x \) is a constant 
\(\displaystyle{ \frac{\partial g}{\partial y} = }\) \(\displaystyle{ \frac{\partial}{\partial y}[x^2y] = }\) \(\displaystyle{ x^2 \frac{\partial}{\partial y}[y] = }\) \(\displaystyle{ x^2 (1) = x^2 }\) 
\(\displaystyle{ \frac{\partial}{\partial y}[x^2y] = x^2 }\) 
Notice that in each case, we could pull out the other variable, since it is considered a constant, and then we take the derivative just as we would in a single variable equation.
For partial derivatives, there are similar rules for products and quotients of functions. Here is a quick video showing those equations.
video by Dr Chris Tisdell 

Before we go on, let's work some practice problems.
Practice
Calculate \(\partial f/ \partial x\) and \(\partial f/ \partial y\) for \(f(x,y) = x^2y + y^3\)
Problem Statement 

Calculate \(\partial f/ \partial x\) and \(\partial f/ \partial y\) for \(f(x,y) = x^2y + y^3\)
Solution 

video by Dr Chris Tisdell 

close solution

Log in to rate this practice problem and to see it's current rating. 

Calculate \( \partial f/ \partial x \), \( \partial f/ \partial y \), \( f_x(0,\pi) \) and \( \partial f(\pi,0)/ \partial y \) for \( f(x,y) = \sin(x)+xy^2 \)
Problem Statement 

Calculate \( \partial f/ \partial x \), \( \partial f/ \partial y \), \( f_x(0,\pi) \) and \( \partial f(\pi,0)/ \partial y \) for \( f(x,y) = \sin(x)+xy^2 \)
Solution 

video by Dr Chris Tisdell 

close solution

Log in to rate this practice problem and to see it's current rating. 

For \( f(x,y) = (x^2yy^3)^5 \), find \( \partial f / \partial x \) and \( \partial f / \partial y \).
Problem Statement 

For \( f(x,y) = (x^2yy^3)^5 \), find \( \partial f / \partial x \) and \( \partial f / \partial y \).
Solution 

video by Krista King Math 

close solution

Log in to rate this practice problem and to see it's current rating. 

Calculate \( \partial z/ \partial x \) and \( \partial z/ \partial y \) for \( z = e^{x^2y} \)
Problem Statement 

Calculate \( \partial z/ \partial x \) and \( \partial z/ \partial y \) for \( z = e^{x^2y} \)
Solution 

video by Dr Chris Tisdell 

close solution

Log in to rate this practice problem and to see it's current rating. 

Chain Rule for Partial Derivatives
These next two videos are very good since they explain the chain rule graphically using some examples. This means you will not have to memorize formulas.
This video looks at the example
(a) \(z=f(x,y), \) \(x=g(t), \) \(y=h(t)\); find \(dz/dt\)
(b) \(z=f(x,y), \) \(x=uv, \) \( y=vu\); find \(z_u\) and \(z_v\)
video by Dr Chris Tisdell 

This video clip looks at
\(w=f(x,y),\) \( x=g(r,s),\) \( y=h(r,s)\); find \( \partial w/ \partial r\) and \(\partial w/ \partial s\)
video by Dr Chris Tisdell 

Here is a very short video clip discussing a more general version of the chain rule for functions of more than two variables.
video by PatrickJMT 

This next video contains a proof for a chain rule for partial derivatives. We recommend that you watch it to get a deeper understanding of the mathematics but it is not required in order to use the chain rule.
This video derives these equations.  
If \( w= f(x) \) and \( x = g(r,s) \) with \(f\) and \(g\) differentiable then  

\(\displaystyle{ \frac{\partial w}{\partial r} = \frac{df}{dx} \frac{\partial g}{\partial r} }\) 
\(\displaystyle{ \frac{\partial w}{\partial s} = \frac{df}{dx} \frac{\partial g}{\partial s} }\) 
video by Dr Chris Tisdell 

Okay, let's practice the chain rule on these problems.
Practice
For \( z = x^2y+xy^2 \), where \( x=2+t^4 \) and \( y = 1t^3 \), calculate \( \partial z / \partial t \).
Problem Statement 

For \( z = x^2y+xy^2 \), where \( x=2+t^4 \) and \( y = 1t^3 \), calculate \( \partial z / \partial t \).
Solution 

video by PatrickJMT 

close solution

Log in to rate this practice problem and to see it's current rating. 

For \( u(x,y) = \sqrt{x^2+y^2} \), where \( x = e^{st} \) and \( y = 1+s^2\cos t \), calculate \( \partial u / \partial t \).
Problem Statement 

For \( u(x,y) = \sqrt{x^2+y^2} \), where \( x = e^{st} \) and \( y = 1+s^2\cos t \), calculate \( \partial u / \partial t \).
Solution 

video by Dr Chris Tisdell 

close solution

Log in to rate this practice problem and to see it's current rating. 

For \( z = x^2 + xy^3 \), where \( x = u v^2 + w^3\) and \( y = u + ve^w\), calculate \( \partial z/ \partial u \), \(\partial z/ \partial v\) and \(\partial z/ \partial w\).
Problem Statement 

For \( z = x^2 + xy^3 \), where \( x = u v^2 + w^3\) and \( y = u + ve^w\), calculate \( \partial z/ \partial u \), \(\partial z/ \partial v\) and \(\partial z/ \partial w\).
Solution 

This solution is shown in two consecutive videos.
video by PatrickJMT 

video by PatrickJMT 

close solution

Log in to rate this practice problem and to see it's current rating. 

First Derivative Test
One of the main uses of the first derivative is in the First Derivative Test, similar to what you learned in single variable calculus. Multivariable functions can be tested with the partial derivative version of the First Derivative Test. This next video shows a proof of how it works. We recommend that you watch this video since it can give you a better feel for partial derivatives.
Although this is a proof, this video is really good to watch since it gives you a feel for how partial derivatives work and what they look like. So don't let the word 'proof' deter you from watching this video. This is one of the best instructors we've ever seen. So he explains it in a way that is very understandable.
video by Dr Chris Tisdell 

Practice
Second Order Partial Derivatives
As you learned in single variable calculus, you can take higher order derivatives of functions. This is also true for multivariable functions. However, for second order partial derivatives, there are actually four second order derivatives, compared to two for single variable functions. Using subscript notation, we have these four partial derivatives.
\(f_{xx}\) 
\(f_{xy}\) 
\(f_{yx}\) 
\(f_{yy}\) 

An important, but not obvious, result is that the two mixed partials, \(f_{xy}\) and \(f_{yx}\) are always equal, i.e. \(f_{xy} = f_{yx}\).
Here is a great video clip explaining these second order equations in more detail using other, very common, notation.
video by Dr Chris Tisdell 

Okay, now let's take the second derivative a step further and use the chain rule. How do we do that? Well, this video shows how and he has a neat way of drawing a diagram to help visualize the chain rule.
video by Dr Chris Tisdell 

Before we go on, let's work some practice problems.
Practice
Calculate all four second order partial derivatives of \( f(x,y) = x^3y + 2 \).
Problem Statement 

Calculate all four second order partial derivatives of \( f(x,y) = x^3y + 2 \).
Solution 

video by Dr Chris Tisdell 

close solution

Log in to rate this practice problem and to see it's current rating. 

For \( h(x,y,z,t) = x^2y \cos(z/t) \), calculate all four first order partial derivatives.
Problem Statement 

For \( h(x,y,z,t) = x^2y \cos(z/t) \), calculate all four first order partial derivatives.
Solution 

video by Krista King Math 

close solution

Log in to rate this practice problem and to see it's current rating. 

For \( f(x,y) = x^2+e^{y^2} \), calculate all four second order partial derivatives.
Problem Statement 

For \( f(x,y) = x^2+e^{y^2} \), calculate all four second order partial derivatives.
Solution 

video by Krista King Math 

close solution

Log in to rate this practice problem and to see it's current rating. 

For \( f(x,y) = x^3y^5 + 2x^4y \), calculate all four second order partial derivatives.
Problem Statement 

For \( f(x,y) = x^3y^5 + 2x^4y \), calculate all four second order partial derivatives.
Solution 

video by Dr Chris Tisdell 

close solution

Log in to rate this practice problem and to see it's current rating. 

Let \( z = \cos(x^2y) \). Calculate \( \partial z/\partial x \) and \( \partial^2z/\partial y \partial x \).
Problem Statement 

Let \( z = \cos(x^2y) \). Calculate \( \partial z/\partial x \) and \( \partial^2z/\partial y \partial x \).
Final Answer 

\( \partial z/\partial x = 2xy\sin(x^2y) \); \( \partial^2z/\partial y \partial x = 2x\sin(x^2y)  2x^3y\cos(x^2y) \)
Problem Statement 

Let \( z = \cos(x^2y) \). Calculate \( \partial z/\partial x \) and \( \partial^2z/\partial y \partial x \).
Solution 

video by Dr Chris Tisdell 

Final Answer 

\( \partial z/\partial x = 2xy\sin(x^2y) \); \( \partial^2z/\partial y \partial x = 2x\sin(x^2y)  2x^3y\cos(x^2y) \) 
close solution

Log in to rate this practice problem and to see it's current rating. 

For \( f(x,y) = ln(x^2+y^2) \), calculate \( f_{xx} \), \( f_{yy} \) and \( f_{xy} \).
Problem Statement 

For \( f(x,y) = ln(x^2+y^2) \), calculate \( f_{xx} \), \( f_{yy} \) and \( f_{xy} \).
Solution 

video by Dr Chris Tisdell 

close solution

Log in to rate this practice problem and to see it's current rating. 

Applications
You will find that one of the major applications of partial derivatives is to solve partial differential equations.
Practice
For \( f(x,y) = 2xy \), calculate \( \partial^2 f / \partial x^2 \) and \( \partial^2 f / \partial y^2 \) and show that \( \partial^2 f / \partial x^2 + \partial^2 f / \partial y^2 = 0 \).
Problem Statement 

For \( f(x,y) = 2xy \), calculate \( \partial^2 f / \partial x^2 \) and \( \partial^2 f / \partial y^2 \) and show that \( \partial^2 f / \partial x^2 + \partial^2 f / \partial y^2 = 0 \).
Solution 

video by Dr Chris Tisdell 

close solution

Log in to rate this practice problem and to see it's current rating. 

Show that the function \( u(x,y) = \ln\sqrt{x^2+y^2} \) satisfies Laplace's equation \( f_{xx} + f_{yy} = 0 \).
Problem Statement 

Show that the function \( u(x,y) = \ln\sqrt{x^2+y^2} \) satisfies Laplace's equation \( f_{xx} + f_{yy} = 0 \).
Solution 

video by Dr Chris Tisdell 

close solution

Log in to rate this practice problem and to see it's current rating. 

Let \( z = f(t) \). If \(\displaystyle{ z = \frac{x+y}{xy} }\), show that \( x^2z_x  y^2z_y = 0 \).
Problem Statement 

Let \( z = f(t) \). If \(\displaystyle{ z = \frac{x+y}{xy} }\), show that \( x^2z_x  y^2z_y = 0 \).
Solution 

video by Dr Chris Tisdell 

close solution

Log in to rate this practice problem and to see it's current rating. 

Let a and b be constants and f is the differentiable function \(w=f(u)\) with \(u=ax+by\). Show that the PDE (Partial Differential Equation) \(aw_ybw_x=0\) holds.
Problem Statement 

Let a and b be constants and f is the differentiable function \(w=f(u)\) with \(u=ax+by\). Show that the PDE (Partial Differential Equation) \(aw_ybw_x=0\) holds.
Solution 

video by Dr Chris Tisdell 

close solution

Log in to rate this practice problem and to see it's current rating. 

Here is a playlist of the videos on this page.
You CAN Ace Calculus
The Unit Circle
The Unit Circle [wikipedia]
Basic Trig Identities
Set 1  basic identities  

\(\displaystyle{ \tan(t) = \frac{\sin(t)}{\cos(t)} }\) 
\(\displaystyle{ \cot(t) = \frac{\cos(t)}{\sin(t)} }\) 
\(\displaystyle{ \sec(t) = \frac{1}{\cos(t)} }\) 
\(\displaystyle{ \csc(t) = \frac{1}{\sin(t)} }\) 
Set 2  squared identities  

\( \sin^2t + \cos^2t = 1\) 
\( 1 + \tan^2t = \sec^2t\) 
\( 1 + \cot^2t = \csc^2t\) 
Set 3  doubleangle formulas  

\( \sin(2t) = 2\sin(t)\cos(t)\) 
\(\displaystyle{ \cos(2t) = \cos^2(t)  \sin^2(t) }\) 
Set 4  halfangle formulas  

\(\displaystyle{ \sin^2(t) = \frac{1\cos(2t)}{2} }\) 
\(\displaystyle{ \cos^2(t) = \frac{1+\cos(2t)}{2} }\) 
Trig Derivatives
\(\displaystyle{ \frac{d[\sin(t)]}{dt} = \cos(t) }\) 
\(\displaystyle{ \frac{d[\cos(t)]}{dt} = \sin(t) }\)  
\(\displaystyle{ \frac{d[\tan(t)]}{dt} = \sec^2(t) }\) 
\(\displaystyle{ \frac{d[\cot(t)]}{dt} = \csc^2(t) }\)  
\(\displaystyle{ \frac{d[\sec(t)]}{dt} = \sec(t)\tan(t) }\) 
\(\displaystyle{ \frac{d[\csc(t)]}{dt} = \csc(t)\cot(t) }\) 
Inverse Trig Derivatives
\(\displaystyle{ \frac{d[\arcsin(t)]}{dt} = \frac{1}{\sqrt{1t^2}} }\) 
\(\displaystyle{ \frac{d[\arccos(t)]}{dt} = \frac{1}{\sqrt{1t^2}} }\)  
\(\displaystyle{ \frac{d[\arctan(t)]}{dt} = \frac{1}{1+t^2} }\) 
\(\displaystyle{ \frac{d[\arccot(t)]}{dt} = \frac{1}{1+t^2} }\)  
\(\displaystyle{ \frac{d[\arcsec(t)]}{dt} = \frac{1}{\abs{t}\sqrt{t^2 1}} }\) 
\(\displaystyle{ \frac{d[\arccsc(t)]}{dt} = \frac{1}{\abs{t}\sqrt{t^2 1}} }\) 
Trig Integrals
\(\int{\sin(x)~dx} = \cos(x)+C\) 
\(\int{\cos(x)~dx} = \sin(x)+C\)  
\(\int{\tan(x)~dx} = \ln\abs{\cos(x)}+C\) 
\(\int{\cot(x)~dx} = \ln\abs{\sin(x)}+C\)  
\(\int{\sec(x)~dx} = \) \( \ln\abs{\sec(x)+\tan(x)}+C\) 
\(\int{\csc(x)~dx} = \) \( \ln\abs{\csc(x)+\cot(x)}+C\) 
To bookmark this page and practice problems, log in to your account or set up a free account.
Single Variable Calculus 

MultiVariable Calculus 

Differential Equations 

Precalculus 

Engineering 

Circuits 

Semiconductors 

Do you have a practice problem number but do not know on which page it is found? If so, enter the number below and click 'page' to go to the page on which it is found or click 'practice' to be taken to the practice problem.
 
The 17Calculus and 17Precalculus iOS and Android apps are no longer available for download. If you are still using a previously downloaded app, your app will be available until the end of 2020, after which the information may no longer be available. However, do not despair. All the information (and more) is now available on 17calculus.com for free. 