You CAN Ace Calculus  

17calculus > partial derivatives  


Topics You Need To Understand For This Page
Calculus Main Topics
Single Variable Calculus 

MultiVariable Calculus 
Tools
math tools 

general learning tools 
additional tools 
Related Topics and Links
external links you may find helpful 

Partial Derivatives 

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.
Dr Chris Tisdell  Product + Quotient rule formulas: Partial derivatives  
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.
Dr Chris Tisdell  (a) \(z=f(x,y), \) \(x=g(t), \) \(y=h(t)\); find \(dz/dt\)  
Dr Chris Tisdell  \(w=f(x,y),\) \( x=g(r,s),\) \( y=h(r,s)\); find \( \partial w/ \partial r\) and \(\partial w/ \partial s\)  
Here is a very short video clip discussing a more general version of the chain rule for functions of more than two variables.
PatrickJMT  General Chain Rule  
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.
Dr Chris Tisdell  Proof of a chain rule for partial derivatives  
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.
Dr Chris Tisdell  Proof: First derivative test  
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.
Dr Chris Tisdell  Partial derivatives: 2nd order examples  
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.
Dr Chris Tisdell  Chain rule: 2nd derivatives example  
Okay, time for some practice problems. Once you are done with those, an important application of partial derivatives can be found on the next page where we discuss gradients and directional derivatives.
next  gradients → 
Search 17Calculus
Practice Problems 

Level A  Basic 
Practice A01  

Find \(\partial f/ \partial x\) and \(\partial f/ \partial y\) for \(f(x,y) = x^2y + y^3\)  
solution 
Practice A02  

Find \( \partial f/ \partial x \), \( \partial f/ \partial y \), \( f_x(0,\pi) \) and \( \partial f(\pi,0)/ \partial y \) for \( f(x,y) = \sin(x)+xy^2 \)  
solution 
Practice A03  

Find all four second order partial derivatives of \( f(x,y) = x^3y+2 \).  
solution 
Practice A04  

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 
Practice A05  

For \( z = x^2y+xy^2 \), where \( x=2+t^4 \) and \( y = 1t^3 \), find \( \partial z / \partial t \).  
solution 
Practice A06  

For \( f(x,y) = (x^2yy^3)^5 \), find \( \partial f / \partial x \) and \( \partial f / \partial y \).  
solution 
Practice A07  

For \( h(x,y,z,t) = x^2y\cos(z/t) \), find all four first order partial derivatives.  
solution 
Practice A08  

For \( f(x,y) = x^2+e^{y^2} \), find all four second order partial derivatives.  
solution 
Practice A09  

For \( f(x,y) = x^3y^5+2x^4y \), find all four second order partial derivatives.  
solution 
Practice A10  

For \( u(x,y)=\sqrt{x^2+y^2} \), where \( x=e^{st} \) and \( y=1+s^2\cos t \), find \( \partial u / \partial t \).  
solution 
Practice A11  

Find \(\partial z/ \partial x\) and \(\partial z/ \partial y\) for \(z=e^{x^2y}\)  
solution 
Practice A12  

Let \(z=\cos(x^2y)\). Calculate \(\partial z/\partial x\) and \(\partial^2z/\partial y \partial x\).  
answer 
solution 
Level B  Intermediate 
Practice B01  

For \(f(x,y)=\ln(x^2+y^2)\), calculate \(f_{xx}\), \(f_{yy}\) and \(f_{xy}\).  
solution 
Practice B02  

For \(z=x^2+xy^3\), where \(x=uv^2+w^3\) and \(y=u+ve^w\), find \(\partial z/ \partial u\), \(\partial z/ \partial v\) and \(\partial z/ \partial w\).  
solution 
Practice B03  

Show that the function \(u(x,y)=\ln\sqrt{x^2+y^2}\) satisfies Laplace's equation \(f_{xx}+f_{yy}=0\).  
solution 
Practice B04  

Let \(z=f(t)\). If \(\displaystyle{z=\frac{x+y}{xy}}\), show that \(x^2z_xy^2z_y=0\).  
solution 
Practice B05  

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