Problem 415
Question
Find the gradient. \(\quad f(x, y)=\frac{\sqrt{x}+y^{2}}{x y}\)
Step-by-Step Solution
Verified Answer
The gradient is \( \nabla f = \left(-\frac{1}{2}x^{-3/2}y^{-1} - x^{-2}y, -x^{-1/2}y^{-2} + x^{-1}\right) \).
1Step 1: Rewrite the Function
First, rewrite the given function to make it easier to differentiate. The function is initially given as \( f(x, y) = \frac{\sqrt{x} + y^2}{xy} \). This can be rewritten as \( f(x, y) = \frac{\sqrt{x}}{xy} + \frac{y^2}{xy} = \frac{x^{1/2}}{xy} + \frac{y}{x} \).
2Step 2: Simplify the Function
Further simplify the rewritten function. It becomes \( f(x, y) = x^{-1/2}y^{-1} + x^{-1}y \).
3Step 3: Partial Derivative with Respect to \(x\)
Find the partial derivative of \( f(x, y) \) with respect to \( x \). Differentiating, we get \( \frac{\partial f}{\partial x} = -\frac{1}{2}x^{-3/2}y^{-1} - x^{-2}y \).
4Step 4: Partial Derivative with Respect to \(y\)
Find the partial derivative of \( f(x, y) \) with respect to \( y \). Differentiating, we obtain \( \frac{\partial f}{\partial y} = -x^{-1/2}y^{-2} + x^{-1} \).
5Step 5: Express the Gradient
The gradient \( abla f \) is a vector consisting of the partial derivatives with respect to \( x \) and \( y \). So, \( abla f = \left(-\frac{1}{2}x^{-3/2}y^{-1} - x^{-2}y, -x^{-1/2}y^{-2} + x^{-1}\right) \).
Key Concepts
Partial DerivativeDifferentiationVector CalculusMultivariable Calculus
Partial Derivative
In the context of multivariable functions, partial derivatives serve as the building blocks for understanding how a function changes with respect to one variable while keeping the others constant. This concept is crucial, especially when dealing with functions of several variables. To compute a partial derivative, we differentiate the function with respect to one variable, treating the other variables as constants.
- For example, if you have a function like \( f(x, y) \), the partial derivative \( \frac{\partial f}{\partial x} \) measures the rate of change of the function with respect to \(x\), assuming \(y\) is constant.
- It gives us insight into the slope of the function along the \(x\) direction.
- Similarly, \( \frac{\partial f}{\partial y} \) measures the change as \(y\) varies.
Differentiation
Differentiation is a mathematical process employed to find the derivative of a function. In simpler terms, it is the method used for assessing how a function's output value changes in response to changes in its input values. Differentiation can be understood more profoundly in the realm of single-variable and multivariable calculus.
In the context of single-variable functions, differentiation involves finding the rate at which a function changes at any given point. However, when extended to functions of multiple variables, such as \(f(x, y)\), differentiation becomes partial differentiation.
In the context of single-variable functions, differentiation involves finding the rate at which a function changes at any given point. However, when extended to functions of multiple variables, such as \(f(x, y)\), differentiation becomes partial differentiation.
- This involves deriving the formula to find the slope of the function surface along the axes of the input variables.
- The differentiation of each variable treats the others as constants, simplifying the path to finding the solution.
- This approach helps in finding critical points, evaluating limits, or understanding the behavior of a system at a specific point.
Vector Calculus
Vector calculus is a branch of mathematics focusing on vector fields and differential and integral operators. It enables handling directional changes, gradients, and transformations in multi-dimensional spaces. This concept becomes especially potent in physics and engineering, where phenomena in 3D space are analyzed.
- The gradient is a vital operator in vector calculus. It points in the direction of the greatest rate of increase of a function and its magnitude denotes the rate of increase.
- For example, in the exercise, the gradient \(abla f\) is represented as a vector combining the partial derivatives of the function with respect to each independent variable.
- Such operations assist in visualizing how a multivariate function behaves and evolves in its domain.
Multivariable Calculus
Multivariable calculus extends the principles of calculus to functions of several variables. Contrary to functions of a single variable, these functions depend on multiple inputs, introducing greater complexity and depth.
- Key operations in multivariable calculus include finding partial derivatives, gradients, and more.
- This branch of calculus allows for the study of optimization, evaluating multiple factors simultaneously, which is particularly beneficial in fields like economics and physical sciences.
- The gradient vector derived in the exercise showcases how multiple partial derivatives can be compiled to understand the direction and rate of change across a multidimensional surface.
Other exercises in this chapter
Problem 412
Find the directional derivative of \(f(x, y)=x^{2}+6 x y-y^{2}\) in the direction \(v=\mathbf{i}+4 \mathbf{j}\).
View solution Problem 413
Find the maximal directional derivative magnitude and direction for the function \(f(x, y)=x^{3}+2 x y-\cos (\pi y)\) at point (3,0).
View solution Problem 416
Find and classify the critical points. \(\quad z=x^{3}-x y+y^{2}-1\)
View solution Problem 417
Use Lagrange multipliers to find the maximum and minimum values for the functions with the given constraints. \(f(x, y)=x^{2} y, x^{2}+y^{2}=4\)
View solution