Explain the meaning of a partial differential equation and give an example. Higher-order partial derivatives can be calculated in the same way as higher-order derivatives. In addition, we will define the gradient vector to help with some of the notation and work here. Then differentiate $$f(x,y)$$ with respect to $$x$$ using the sum, difference, and power rules: \begin{align*}\dfrac{∂f}{∂x} &=\dfrac{∂}{∂x}\left[x^2−3xy+2y^2−4x+5y−12\right] \\[6pt] &=\dfrac{∂}{∂x}[x^2]−\dfrac{∂}{∂x}[3xy]+\dfrac{∂}{∂x}[2y^2]−\dfrac{∂}{∂x}[4x]+\dfrac{∂}{∂x}[5y]−\dfrac{∂}{∂x}[12] \\[6pt] &=2x−3y+0−4+0−0 \\ &=2x−3y−4. Since there isn’t too much to this one, we will simply give the derivatives. A partial differential equation is an equation that involves an unknown function of more than one independent variable and one or more of its partial derivatives. So, the partial derivatives from above will more commonly be written as. Today’s accepted value of Earth’s age is about 4.6 billion years. Second partial derivatives (one seems strange) 5. &=−\dfrac{\sqrt{5}}{2} ≈−1.118. This means the third term will differentiate to zero since it contains only $$x$$’s while the $$x$$’s in the first term and the $$z$$’s in the second term will be treated as multiplicative constants. To calculate $$\dfrac{∂^2f}{∂x∂y}$$ and $$\dfrac{∂^2f}{∂y^2}$$, first calculate $$∂f/∂y$$: \[\dfrac{∂f}{∂y}=−3xe^{−3y}−5\cos(2x−5y). Let’s start out by differentiating with respect to $$x$$. Two other second-order partial derivatives can be calculated for any function $$f(x,y).$$ The partial derivative $$f_{xx}$$ is equal to the partial derivative of $$f_x$$ with respect to $$x$$, and $$f_{yy}$$ is equal to the partial derivative of $$f_y$$ with respect to $$y$$. Just as with derivatives of single-variable functions, we can call these second-order derivatives, third-order derivatives, and so on. For a function of two variables, and are the independent variables and is the dependent variable. &=x^2\cos(x^2y−z)+z\sin(x^2−yz) \end{align*}, \begin{align*} \dfrac{∂f}{∂z} &=\dfrac{∂}{∂z}[\sin(x^2y−z)+\cos(x^2−yz)] \\[6pt] (b) A close-up of the results at a depth of, Next: Tangent Planes and Linear Approximations, Creative Commons Attribution-NonCommercial-ShareAlike 4.0 International License. This is important because we are going to treat all other variables as constants and then proceed with the derivative as if it was a function of a single variable. Before we work any examples let’s get the formal definition of the partial derivative out of the way as well as some alternate notation. For example, if we have a function of and we wish to calculate then we treat the other two independent variables as if they are constants, then differentiate with respect to, Use the limit definition of partial derivatives to calculate for the function. Find and when and, Suppose the sides of a rectangle are changing with respect to time. Remember how to differentiate natural logarithms. &=0−0+0−4x+10yz−0+0−3 \\[4pt] In this case we treat all $$x$$’s as constants and so the first term involves only $$x$$’s and so will differentiate to zero, just as the third term will. Here are the formal definitions of the two partial derivatives we looked at above. \end{align*}. You can specify any order of integration. A Cobb-Douglas production function is where represent the amount of labor and capital available. To calculate $$∂g/∂x,$$ treat the variable y as a constant. Calculate the three partial derivatives of the following functions. The area of a parallelogram with adjacent side lengths that are and in which the angle between these two sides is is given by the function Find the rate of change of the area of the parallelogram with respect to the following: Express the volume of a right circular cylinder as a function of two variables: Find the indicated higher-order partial derivatives. His conclusion was a range of 20 to 400 million years, but most likely about 50 million years. To calculate $$\dfrac{∂^2f}{∂y^2}$$, differentiate $$∂f/∂y$$ (Equation \ref{Ex6e5}) with respect to $$y$$: \begin{align*} \dfrac{∂^2f}{∂y^2} &=\dfrac{∂}{∂y}\left[\dfrac{∂f}{∂y}\right] \\[6pt] A graph of this solution using appears in (Figure), where the initial temperature distribution over a wire of length is given by Notice that as time progresses, the wire cools off. We’ll do the same thing for this function as we did in the previous part. In each case, treat all variables as constants except the one whose partial derivative you are calculating. \end{align*}. \end{align*}\], To calculate $$\dfrac{∂f}{∂y}$$, first calculate $$f(x,y+h):$$, \begin{align*} f(x+h,y) &=x^2−3x(y+h)+2(y+h)^2−4x+5(y+h)−12 \\ &=x^2−3xy−3xh+2y^2+4yh+2h^2−4x+5y+5h−12. For the partial derivative with respect to h we hold r constant: f’ h = π r 2 (1)= π r 2 (π and r 2 are constants, and the derivative of h with respect to h is 1) It says "as only the height changes (by the tiniest amount), the volume changes by π r 2 " It is like we add the thinnest disk on top with a circle's area of π r 2. You appear to be on a device with a "narrow" screen width (, Derivatives of Exponential and Logarithm Functions, L'Hospital's Rule and Indeterminate Forms, Substitution Rule for Indefinite Integrals, Volumes of Solids of Revolution / Method of Rings, Volumes of Solids of Revolution/Method of Cylinders, Parametric Equations and Polar Coordinates, Gradient Vector, Tangent Planes and Normal Lines, Triple Integrals in Cylindrical Coordinates, Triple Integrals in Spherical Coordinates, Linear Homogeneous Differential Equations, Periodic Functions & Orthogonal Functions, Heat Equation with Non-Zero Temperature Boundaries, Absolute Value Equations and Inequalities. Remember how to differentiate natural logarithms. &=\dfrac{∂}{∂y}[−3xe^{−3y}−5\cos(2x−5y)] \\[6pt] Figure $$\PageIndex{2}$$ represents a contour map for the function $$g(x,y)$$. Get the free "Partial Derivative Calculator" widget for your website, blog, Wordpress, Blogger, or iGoogle. So, there are some examples of partial derivatives. We will see an easier way to do implicit differentiation in a later section. Then, \[\begin{align*} h′(x) &=\lim_{k→0}\dfrac{h(x+k)−h(x)}{k} \\[6pt] &=\lim_{k→0}\dfrac{f(x,y+k)−f(x,y)}{k} \\[6pt] &=\dfrac{∂f}{∂y}. For instance, one variable could be changing faster than the other variable(s) in the function. The discovery of radioactivity came near the end of Kelvin’s life and he acknowledged that his calculation would have to be modified. Note as well that we usually don’t use the $$\left( {a,b} \right)$$ notation for partial derivatives as that implies we are working with a specific point which we usually are not doing. because we are now working with functions of multiple variables. The heat equation in one dimension becomes, where $$c^2$$ represents the thermal diffusivity of the material in question. If the functions $$f_{xy}$$ and $$f_{yx}$$ are continuous on $$D$$, then $$f_{xy}=f_{yx}$$. We will need to develop ways, and notations, for dealing with all of these cases. This means that the second and fourth terms will differentiate to zero since they only involve $$y$$’s and $$z$$’s. His conclusion was a range of million years, but most likely about million years. Now let’s solve for $$\frac{{\partial z}}{{\partial x}}$$. Kelvin made reasonable assumptions based on what was known in his time, but he also made several assumptions that turned out to be wrong. That prophetic utterance referred to what we are now considering tonight, radium! Now, we can verify through direct substitution for each equation that the solutions are $$f(t)=Ae^{−λ^2t}$$ and $$R(r)=B\left(\dfrac{\sin αr}{r}\right)+C\left(\dfrac{\cos αr}{r}\right)$$, where $$α=λ/\sqrt{K}$$. Limits – In the section we’ll take a quick look at evaluating limits of functions of several variables. Therefore, represents the slope of the tangent line passing through the point parallel to the and represents the slope of the tangent line passing through the point parallel to the If we wish to find the slope of a tangent line passing through the same point in any other direction, then we need what are called directional derivatives, which we discuss in Directional Derivatives and the Gradient. Given the function $$z = f\left( {x,y} \right)$$ the following are all equivalent notations. This one will be slightly easier than the first one. \dfrac{∂^2f}{∂x∂y} &=\dfrac{∂}{∂x}\left[\dfrac{∂f}{∂y}\right] \\[4pt] For the following exercises, calculate the sign of the partial derivative using the graph of the surface. &=(\cos(x^2y−z))\dfrac{∂}{∂x}(x^2y−z)−(\sin(x^2−yz))\dfrac{∂}{∂x}(x^2−yz) \\[6pt] \nonumber, \[\begin{align*} f(x+h,y) &=(x+h)^2−3(x+h)y+2y^2−4(x+h)+5y−12 \\ &=x^2+2xh+h^2−3xy−3hy+2y^2−4x−4h+5y−12. Today’s accepted value of Earth’s age is about billion years. During the late 1800s, the scientists of the new field of geology were coming to the conclusion that Earth must be “millions and millions” of years old. For more information contact us at info@libretexts.org or check out our status page at https://status.libretexts.org.