17Calculus - Partial Derivatives
Partial derivatives follow directly from derivatives you have seen in single variable calculus. Calculation is pretty straightforward but, as is common in multi-variable calculus, you need to watch your notation carefully.
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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 | multi-variable | |
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\(\displaystyle{ \frac{df}{dx} }\) 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 \(x\) \(\left[ \partial g/\partial x \right]\) |
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\( y \) 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 \(y\) \(\left[ \partial g/\partial y \right]\) |
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\( x \) 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 [51secs]
video by Dr Chris Tisdell |
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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 |
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Calculate \(\partial f/ \partial x\) and \(\partial f/ \partial y\) for \(f(x,y) = x^2y + y^3\)
Solution |
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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 |
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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 |
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For \( f(x,y) = (x^2y-y^3)^5 \), find \( \partial f / \partial x \) and \( \partial f / \partial y \).
Problem Statement |
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For \( f(x,y) = (x^2y-y^3)^5 \), find \( \partial f / \partial x \) and \( \partial f / \partial y \).
Solution |
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Calculate \( \partial z/ \partial x \) and \( \partial z/ \partial y \) for \( z = e^{x^2y} \)
Problem Statement |
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Calculate \( \partial z/ \partial x \) and \( \partial z/ \partial y \) for \( z = e^{x^2y} \)
Solution |
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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) \(z=f(x,y), \) \(x=u-v, \) \( y=v-u\); find \(z_u\) and \(z_v\)
Dr Chris Tisdell - Chain rule example: functions of two variables [6mins-34secs]
video by Dr Chris Tisdell |
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This video clip looks at
\(w=f(x,y),\) \( x=g(r,s),\) \( y=h(r,s)\); find \( \partial w/ \partial r\) and \(\partial w/ \partial s\)
Dr Chris Tisdell - Chain rule and Partial Differential Equations [55secs]
video by Dr Chris Tisdell |
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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 [41secs]
video by PatrickJMT |
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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 [11mins-5secs]
This video derives these equations. | ||
If \( w= f(x) \) and \( x = g(r,s) \) with \(f\) and \(g\) differentiable then | ||
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\(\displaystyle{ \frac{\partial w}{\partial r} = \frac{df}{dx} \frac{\partial g}{\partial r} }\) |
\(\displaystyle{ \frac{\partial w}{\partial s} = \frac{df}{dx} \frac{\partial g}{\partial s} }\) | |
video by Dr Chris Tisdell |
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Okay, let's practice the chain rule on these problems.
Practice
For \( z = x^2y+xy^2 \), where \( x=2+t^4 \) and \( y = 1-t^3 \), calculate \( \partial z / \partial t \).
Problem Statement |
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For \( z = x^2y+xy^2 \), where \( x=2+t^4 \) and \( y = 1-t^3 \), calculate \( \partial z / \partial t \).
Solution |
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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 |
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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 |
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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 |
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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 |
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This solution is shown in two consecutive videos.
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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. Multi-variable 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 [9mins-6secs]
video by Dr Chris Tisdell |
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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 multi-variable 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.
\(f_{xx}\) |
\(f_{xy}\) |
\(f_{yx}\) |
\(f_{yy}\) |
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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 [2mins]
video by Dr Chris Tisdell |
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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 [5mins-10secs]
video by Dr Chris Tisdell |
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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 |
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Calculate all four second order partial derivatives of \( f(x,y) = x^3y + 2 \).
Solution |
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For \( h(x,y,z,t) = x^2y \cos(z/t) \), calculate all four first order partial derivatives.
Problem Statement |
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For \( h(x,y,z,t) = x^2y \cos(z/t) \), calculate all four first order partial derivatives.
Solution |
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For \( f(x,y) = x^2+e^{y^2} \), calculate all four second order partial derivatives.
Problem Statement |
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For \( f(x,y) = x^2+e^{y^2} \), calculate all four second order partial derivatives.
Solution |
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For \( f(x,y) = x^3y^5 + 2x^4y \), calculate all four second order partial derivatives.
Problem Statement |
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For \( f(x,y) = x^3y^5 + 2x^4y \), calculate all four second order partial derivatives.
Solution |
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Let \( z = \cos(x^2y) \). Calculate \( \partial z/\partial x \) and \( \partial^2z/\partial y \partial x \).
Problem Statement |
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Let \( z = \cos(x^2y) \). Calculate \( \partial z/\partial x \) and \( \partial^2z/\partial y \partial x \).
Final Answer |
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\( \partial z/\partial x = -2xy\sin(x^2y) \); \( \partial^2z/\partial y \partial x = -2x\sin(x^2y) - 2x^3y\cos(x^2y) \)
Problem Statement |
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Let \( z = \cos(x^2y) \). Calculate \( \partial z/\partial x \) and \( \partial^2z/\partial y \partial x \).
Solution |
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Final Answer |
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\( \partial z/\partial x = -2xy\sin(x^2y) \); \( \partial^2z/\partial y \partial x = -2x\sin(x^2y) - 2x^3y\cos(x^2y) \) |
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For \( f(x,y) = ln(x^2+y^2) \), calculate \( f_{xx} \), \( f_{yy} \) and \( f_{xy} \).
Problem Statement |
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For \( f(x,y) = ln(x^2+y^2) \), calculate \( f_{xx} \), \( f_{yy} \) and \( f_{xy} \).
Solution |
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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 |
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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 |
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Show that the function \( u(x,y) = \ln\sqrt{x^2+y^2} \) satisfies Laplace's equation \( f_{xx} + f_{yy} = 0 \).
Problem Statement |
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Show that the function \( u(x,y) = \ln\sqrt{x^2+y^2} \) satisfies Laplace's equation \( f_{xx} + f_{yy} = 0 \).
Solution |
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Let \( z = f(t) \). If \(\displaystyle{ z = \frac{x+y}{xy} }\), show that \( x^2z_x - y^2z_y = 0 \).
Problem Statement |
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Let \( z = f(t) \). If \(\displaystyle{ z = \frac{x+y}{xy} }\), show that \( x^2z_x - y^2z_y = 0 \).
Solution |
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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) \(aw_y-bw_x=0\) holds.
Problem Statement |
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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) \(aw_y-bw_x=0\) holds.
Solution |
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partial derivatives 17calculus youtube playlist
Here is a playlist of the videos on this page.
You CAN Ace Calculus
Topics You Need To Understand For This Page
Trig Formulas
The Unit Circle
The Unit Circle [wikipedia]
Basic Trig Identities
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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 | ||
|---|---|---|
\( \sin^2t + \cos^2t = 1\) |
\( 1 + \tan^2t = \sec^2t\) |
\( 1 + \cot^2t = \csc^2t\) |
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Set 3 - double-angle formulas | |
|---|---|
\( \sin(2t) = 2\sin(t)\cos(t)\) |
\(\displaystyle{ \cos(2t) = \cos^2(t) - \sin^2(t) }\) |
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Set 4 - half-angle formulas | |
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\(\displaystyle{ \sin^2(t) = \frac{1-\cos(2t)}{2} }\) |
\(\displaystyle{ \cos^2(t) = \frac{1+\cos(2t)}{2} }\) |
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
\(\displaystyle{ \frac{d[\arcsin(t)]}{dt} = \frac{1}{\sqrt{1-t^2}} }\) |
\(\displaystyle{ \frac{d[\arccos(t)]}{dt} = -\frac{1}{\sqrt{1-t^2}} }\) | |
\(\displaystyle{ \frac{d[\arctan(t)]}{dt} = \frac{1}{1+t^2} }\) |
\(\displaystyle{ \frac{d[\arccot(t)]}{dt} = -\frac{1}{1+t^2} }\) | |
\(\displaystyle{ \frac{d[\arcsec(t)]}{dt} = \frac{1}{\abs{t}\sqrt{t^2 -1}} }\) |
\(\displaystyle{ \frac{d[\arccsc(t)]}{dt} = -\frac{1}{\abs{t}\sqrt{t^2 -1}} }\) |
Trig Integrals
\(\int{\sin(x)~dx} = -\cos(x)+C\) |
\(\int{\cos(x)~dx} = \sin(x)+C\) | |
\(\int{\tan(x)~dx} = -\ln\abs{\cos(x)}+C\) |
\(\int{\cot(x)~dx} = \ln\abs{\sin(x)}+C\) | |
\(\int{\sec(x)~dx} = \) \( \ln\abs{\sec(x)+\tan(x)}+C\) |
\(\int{\csc(x)~dx} = \) \( -\ln\abs{\csc(x)+\cot(x)}+C\) |
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Differential Equations |
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Precalculus |
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