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- https://math.libretexts.org/Bookshelves/Calculus/Map%3A_Calculus__Early_Transcendentals_(Stewart)/14%3A_Partial_Derivatives/14.06%3A_Directional_Derivatives_and_the_Gradient_Vector\[\begin{align*} \vecs ∇f(x,y,z) &=f_x(x,y,z)\,\hat{\mathbf i}+f_y(x,y,z)\,\hat{\mathbf j}+f_z(x,y,z)\,\hat{\mathbf k} \\ &=(2e^{−2z}\cos 2x \cos 2y)\,\hat{\mathbf i}+(−2e^{−2z} \sin 2x \sin 2y)\,\hat...\[\begin{align*} \vecs ∇f(x,y,z) &=f_x(x,y,z)\,\hat{\mathbf i}+f_y(x,y,z)\,\hat{\mathbf j}+f_z(x,y,z)\,\hat{\mathbf k} \\ &=(2e^{−2z}\cos 2x \cos 2y)\,\hat{\mathbf i}+(−2e^{−2z} \sin 2x \sin 2y)\,\hat{\mathbf j}+(−2e^{−2z}\sin 2x \cos 2y)\,\hat{\mathbf k} \\ &=2e^{−2z}(\cos 2x \cos 2y \,\hat{\mathbf i}−\sin 2x \sin 2y\,\hat{\mathbf j}−\sin 2x \cos 2y\,\hat{\mathbf k}). \end{align*}\]
- https://math.libretexts.org/Bookshelves/Calculus/Vector_Calculus_(Corral)/02%3A_Functions_of_Several_Variables/2.04%3A_Directional_Derivatives_and_the_GradientFor a function z=f(x,y), we learned that the partial derivatives ∂f/∂x and∂f/∂y represent the (instantaneous) rate of change of f in the positive x and y directions, respectively. What about other dir...For a function z=f(x,y), we learned that the partial derivatives ∂f/∂x and∂f/∂y represent the (instantaneous) rate of change of f in the positive x and y directions, respectively. What about other directions? It turns out that we can find the rate of change in any direction using a more general type of derivative called a directional derivative.
- https://math.libretexts.org/Courses/De_Anza_College/Calculus_IV%3A_Multivariable_Calculus/01%3A_Differentiation_of_Functions_of_Several_Variables/1.06%3A_Directional_Derivatives_and_the_GradientA function \(z=f(x,y)\) has two partial derivatives: \(∂z/∂x\) and \(∂z/∂y\). These derivatives correspond to each of the independent variables and can be interpreted as instantaneous rates of change ...A function \(z=f(x,y)\) has two partial derivatives: \(∂z/∂x\) and \(∂z/∂y\). These derivatives correspond to each of the independent variables and can be interpreted as instantaneous rates of change (that is, as slopes of a tangent line). Similarly, \(∂z/∂y\) represents the slope of the tangent line parallel to the y-axis. Now we consider the possibility of a tangent line parallel to neither axis.
- https://math.libretexts.org/Courses/Misericordia_University/MTH_171-172%3A_Calculus_-_Early_Transcendentals_(Stewart)/14%3A_Partial_Derivatives/14.06%3A_Directional_Derivatives_and_the_Gradient_Vector\[\begin{align*} \vecs ∇f(x,y,z) &=f_x(x,y,z)\,\hat{\mathbf i}+f_y(x,y,z)\,\hat{\mathbf j}+f_z(x,y,z)\,\hat{\mathbf k} \\ &=(2e^{−2z}\cos 2x \cos 2y)\,\hat{\mathbf i}+(−2e^{−2z} \sin 2x \sin 2y)\,\hat...\[\begin{align*} \vecs ∇f(x,y,z) &=f_x(x,y,z)\,\hat{\mathbf i}+f_y(x,y,z)\,\hat{\mathbf j}+f_z(x,y,z)\,\hat{\mathbf k} \\ &=(2e^{−2z}\cos 2x \cos 2y)\,\hat{\mathbf i}+(−2e^{−2z} \sin 2x \sin 2y)\,\hat{\mathbf j}+(−2e^{−2z}\sin 2x \cos 2y)\,\hat{\mathbf k} \\ &=2e^{−2z}(\cos 2x \cos 2y \,\hat{\mathbf i}−\sin 2x \sin 2y\,\hat{\mathbf j}−\sin 2x \cos 2y\,\hat{\mathbf k}). \end{align*}\]
- https://math.libretexts.org/Courses/Irvine_Valley_College/Math_4A%3A_Multivariable_Calculus/02%3A_Functions_of_Several_Variables/2.04%3A_More_on_Multivariable_DerivativesIn single-variable calculus, we found that one of the most useful differentiation rules is the chain rule, which allows us to find the derivative of the composition of two functions. We follow this wi...In single-variable calculus, we found that one of the most useful differentiation rules is the chain rule, which allows us to find the derivative of the composition of two functions. We follow this with a discussion on the directional derivative, which is a derivative taken in a particular direction along a surface (instead of the two default directions which are parallel to the \(x\) and \(y\) axis, yielding the classic partial derivatives.)
- https://math.libretexts.org/Bookshelves/Scientific_Computing_Simulations_and_Modeling/Introduction_to_the_Modeling_and_Analysis_of_Complex_Systems_(Sayama)/13%3A_Continuous_Field_Models_I__Modeling/13.02%3A_Fundamentals_of_Vector_CalculusIn order to develop continuous field models, you need to know some basic mathematical concepts developed and used in vector calculus.
