The excitement of learning separates youth from old age. As long as you’re learning, you’re not old, Rosalyn S.Yalow.
A function of two variables f: ℝ x ℝ → ℝ assigns to each ordered pair in its domain a unique real number, e.g., Area = $\frac{1}{2}b·h$, z = f(x, y) = 2x + 3y, f(x, y) = x^{2} + y^{2}, e^{x+y}, etc.
Partial derivatives are derivatives of a function of multiple variables, say f(x_{1}, x_{2}, ···, x_{n}) with respect to one of those variables, with the other variables held constant. They measure how the function changes as one variable changes, while keeping the others constant. $\frac{\partial f}{\partial x}(x_0, y_0) = \lim_{\Delta x \to 0} \frac{f(x_0+\Delta x, y_0)-f(x_0, y_0)}{\Delta x}$
Geometrically, the partial derivative $\frac{\partial f}{\partial x}(x_0, y_0)$ at a point (x_{0}, y_{0}) can be interpreted as the slope of the tangent line to the curve formed by the intersection of the surface defined by f(x,y) and the plane y = y_{0}. Similarly, $\frac{\partial f}{\partial y}(x_0, y_0)$ represents the slope of the tangent line to the curve formed by the intersection of the surface defined by f(x,y) and the plane x = x_{0}
$\frac{dw}{ds}\bigg|_{\vec{u}} = ∇w·\vec{u} = |∇w|·|\vec{u}|cos(θ) = |∇w|·cos(θ)$ where θ is the angle between the gradient and the given unit vector.
The gradient is the direction in which the function increases fastest (the direction of the steepest ascent) at a given point, and |∇w| = $\frac{dw}{ds}\bigg|_{\vec{u}=dir(∇w)}$.
The directional derivative is minimized when cos(θ) = -1 ↭ θ = 180 ↭ $\vec{u}$ is in the opposite direction of the gradient ∇w. Futhermore, $\frac{dw}{ds}\bigg|_{\vec{u}} = 0$ ↭ cos(θ) = 0 ↭ θ = 90° ↭ $\vec{u}$ ⊥ ∇w.
The Chain Rule. Let w = f(x, y, z) where x = x(t), y = y(t), and z = z(t) are functions of another variable t, then the derivate of f with respect to t is given by $\frac{dw}{dt} = f_x\frac{dx}{dt} +f_y\frac{dy}{dt}+f_z\frac{dz}{dt}, \frac{dw}{dt} = w_x\frac{dx}{dt} +w_y\frac{dy}{dt}+w_z\frac{dz}{dt} = ∇w·\frac{d\vec{r}}{dt}~$ where ∇w = ⟨w_{x}, w_{y}, w_{z}⟩ and $\frac{d\vec{r}}{dt}=⟨\frac{d\vec{x}}{dt}, \frac{d\vec{y}}{dt}, \frac{d\vec{z}}{dt}⟩$
The gradient vector is defined as follows: ∇w = ⟨w_{x}, w_{y}, w_{z}⟩ = $⟨\frac{\partial w}{\partial x}, \frac{\partial w}{\partial y}, \frac{\partial w}{\partial z}⟩$. And $\frac{d\vec{r}}{dt}$ is the vector of derivatives of x, y, and z with respect to t: $\frac{d\vec{r}}{dt} = ⟨\frac{dx}{dt}, \frac{dy}{dt}, \frac{dz}{dt}⟩$.
The gradient vector is a vector that points in the direction of the steepest increase of the function at a given point. It is perpendicular (orthogonal) to the level surfaces of the function. This means that the gradient points in the direction of the steepest ascent or increase, and there is no change in the function’s value along the level surface, making the gradient perpendicular to it.
In mathematics, some equations involving two variables, x and y, do not explicitly define y as a function of x. This means that even though y can be expressed as a function of x implicitly, the equation might be too complex or impossible to rearrange in a way that solves for y explicitly in terms of x. These equations are said to be defined implicitly.
For example, consider an equation like x^{2} + y^{2} = 1. This equation implicitly defines a relationship between x and y, but it does not express y directly as a function of x (such as y = f(x)). However, there might still exist a function y=f(x) that satisfies this equation for each x.
The technique of implicit differentiation allows us to find the derivative of y with respect to x (denoted as $\frac{dy}{dx}$) without needing to solve the equation explicitly for y.
Examples:
Now, differentiate both sides with respect to x: $\frac{d}{dx}sin(y)=1⇨~ cos(y)\frac{dy}{dx}=1⇨~\frac{dy}{dx} = \frac{1}{cos(y)}$
$y’= \frac{1}{cos(y)}=~ \frac{1}{\sqrt{1-sin^{2}y}}=~ \frac{1}{\sqrt{1-x^{2}}}$
$(sin^{-1}x)’=\frac{d}{dx}sin^{-1}x=~ \frac{1}{\sqrt{1-x^{2}}}$
Rewrite the Implicit Equation. First, let’s rewrite the equation in a form that’s easier to differentiate: f(x, y, z) = $x^2+y^4-z^3+3xy^2-8 = 0$. It defines z implicitly as a function of x and y.
Compute the Partial derivatives of f(x, y, z)
Next, we need to calculate the partial derivatives of f(x, y, z) with respect to x, y, and z respectively:
$f_x = \frac{∂f}{∂x} = 2x + 3y^2, f_y = \frac{∂f}{∂y} = 4y^3+6xy, f_z = \frac{∂f}{∂z} = -3z^2$
Implicit Differentiation.
To compute $\frac{∂z}{∂x}, \frac{∂z}{∂y}$, we use implicit differentiation. The key idea is to differentiate the implicit equation f(x, y, z) = 0 with respect to the desired variable while treating z as a function of x and y, z = z(x, y).
Partial derivative $\frac{∂z}{∂x}$
Differentiate f(x, y, z) = 0 with respect to x: $\frac{∂f}{∂x}\frac{∂x}{∂x} + \frac{∂f}{∂y}\frac{∂y}{∂x}+ \frac{∂f}{∂z}\frac{∂z}{∂x}=$
Since y is fixed, we have: $\frac{∂x}{∂x}=1, \frac{∂y}{∂x} = 0$.
So, the previous equation simplifies to: $\frac{∂f}{∂x}+\frac{∂f}{∂z}\frac{∂z}{∂x} = f_x+f_z\frac{∂z}{∂x}=0⇒[\text{Solve for} \frac{∂z}{∂x}] \frac{∂z}{∂x}=z_x = \frac{-f_x}{f_z}$
$\frac{∂z}{∂x}= z_x = \frac{-f_x}{f_z} =[\text{Plug in the values}] \frac{-(2x+3y^2)}{-3z^2} = \frac{2x+3y^2}{3z^2}$.
Partial derivative $\frac{∂z}{∂y}$
Similarly, to find $\frac{∂z}{∂y}$, we differentiate f(x, y, z) = 0 with respect to y:
$\frac{∂f}{∂x}\frac{∂x}{∂y} + \frac{∂f}{∂y}\frac{∂y}{∂y}+ \frac{∂f}{∂z}\frac{∂z}{∂y}=$
Since x is fixed, we have: $\frac{∂x}{∂y}=0, \frac{∂y}{∂y} = 1$
So, the equation simplifies to: $f_x·0 + f_y·1 + f_z·\frac{∂z}{∂y} = 0↭ f_y + f_z·\frac{∂z}{∂y} = 0$. Solve for $\frac{∂z}{∂y}$: $\frac{∂z}{∂y} = \frac{-f_y}{f_z}$
$\frac{∂z}{∂y}= z_y = \frac{-f_y}{f_z} =[\text{Plug in the values}] \frac{-( 4y^3+6xy)}{-3z^2} = \frac{4y^3+6xy}{3z^2}$·
Rewrite the Implicit Equation. First, let’s rewrite the equation in a form that’s easier to differentiate: f(x, y, z) = x^{2}sin(2y -5z) - 1 - ycos(6zx) = 0$. It defines z implicitly as a function of x and y.
Compute the Partial derivatives of f(x, y, z)
Next, we need to calculate the partial derivatives of f(x, y, z) with respect to x, y, and z respectively:
$f_x = \frac{∂f}{∂x} = 2xsin(2y -5z) +6yzsin(6zx), f_y = \frac{∂f}{∂y} = 2x^2cos(2y -5z) -cos(6zx) , f_z = \frac{∂f}{∂z} = -5x^2cos(2y -5z) +6yxsin(6zx)$
Implicit Differentiation.
To compute $\frac{∂z}{∂x}, \frac{∂z}{∂y}$, we use implicit differentiation. The key idea is to differentiate the implicit equation f(x, y, z) = 0 with respect to the desired variable while treating z as a function of x and y, z = z(x, y).
Partial derivative $\frac{∂z}{∂x}$
Differentiate f(x, y, z) = 0 with respect to x: $\frac{∂f}{∂x}\frac{∂x}{∂x} + \frac{∂f}{∂y}\frac{∂y}{∂x}+ \frac{∂f}{∂z}\frac{∂z}{∂x}=$
Since y is fixed, we have: $\frac{∂x}{∂x}=1, \frac{∂y}{∂x} = 0$.
So, the previous equation simplifies to: $\frac{∂f}{∂x}+\frac{∂f}{∂z}\frac{∂z}{∂x} = f_x+f_z\frac{∂z}{∂x}=0⇒[\text{Solve for} \frac{∂z}{∂x}] \frac{∂z}{∂x}=z_x = \frac{-f_x}{f_z}$
$\frac{∂z}{∂x}= z_x = \frac{-f_x}{f_z} =[\text{Plug in the values}] \frac{-(2xsin(2y -5z) +6yzsin(6zx))}{-5x^2cos(2y -5z) +6yxsin(6zx)} = \frac{2xsin(2y -5z) +6yzsin(6zx)}{5x^2cos(2y -5z) -6yxsin(6zx)}$.
Partial derivative $\frac{∂z}{∂y}$
Similarly, to find $\frac{∂z}{∂y}$, we differentiate f(x, y, z) = 0 with respect to y:
$\frac{∂f}{∂x}\frac{∂x}{∂y} + \frac{∂f}{∂y}\frac{∂y}{∂y}+ \frac{∂f}{∂z}\frac{∂z}{∂y}=$
Since x is fixed, we have: $\frac{∂x}{∂y}=0, \frac{∂y}{∂y} = 1$
So, the equation simplifies to: $f_x·0 + f_y·1 + f_z·\frac{∂z}{∂y} = 0↭ f_y + f_z·\frac{∂z}{∂y} = 0$. Solve for $\frac{∂z}{∂y}$: $\frac{∂z}{∂y} = \frac{-f_y}{f_z}$
$\frac{∂z}{∂y}= z_y = \frac{-f_y}{f_z} =[\text{Plug in the values}] \frac{-(2x^2cos(2y -5z) -cos(6zx))}{-5x^2cos(2y -5z) +6yxsin(6zx)} = \frac{2x^2cos(2y -5z)-cos(6zx)}{5x^2cos(2y -5z) -6yxsin(6zx)}$·
Rewrite the Implicit Equation. It is not necessarily.
Compute the Partial derivatives of f(x, y, z)
Next, we need to calculate the partial derivatives of f(x, y, z) with respect to x, y, and z respectively:
$f_x = \frac{∂f}{∂x} = 4x^3+2xy^2z^2, f_y = \frac{∂f}{∂y} = 4y^3 + 2yx^2z^2, f_z = \frac{∂f}{∂z} = 4z^3+2zx^2y^2$
Implicit Differentiation.
To compute $\frac{∂y}{∂z}$, we use implicit differentiation. The key idea is to differentiate the implicit equation f(x, y, z) = 0 with respect to the desired variable while treating y as a function of x and x, z = y(x, z).
Partial derivative $\frac{∂y}{∂z}$
Differentiate f(x, y, z) = 0 with respect to z: $\frac{∂f}{∂x}\frac{∂x}{∂z} + \frac{∂f}{∂y}\frac{∂y}{∂z}+ \frac{∂f}{∂z}\frac{∂z}{∂z}=$
Since x is fixed, we have: $\frac{∂x}{∂z}=0, \frac{∂z}{∂z} = 1$.
So, the previous equation simplifies to: $\frac{∂f}{∂y}\frac{∂y}{∂z}+ \frac{∂f}{∂z} = f_y\frac{∂y}{∂z}+f_z=0⇒[\text{Solve for} \frac{∂y}{∂z}] \frac{∂y}{∂z}=y_z = \frac{-f_z}{f_y}$
$y_z = \frac{-f_z}{f_y} =[\text{Plug in the values}] \frac{-(4z^3+2zx^2y^2)}{4y^3 + 2yx^2z^2} = -\frac{z(2z^2+x^2y^2)}{y(2y^2 + x^2z^2)}$.