Problem 19 - Calculus for Biology and Medicine

Question

Find the gradient of each function. $$ f(x, y)=\sqrt{x^{3}-3 x y} $$

Step-by-Step Solution

Answer

\(\nabla f(x, y) = \left( \frac{3x^2 - 3y}{2\sqrt{x^3 - 3xy}}, \frac{-3x}{2\sqrt{x^3 - 3xy}} \right)\)

Step 1: Understanding the Gradient

The gradient of a function in two variables, such as \(f(x, y)\), is a vector that consists of its partial derivatives with respect to each variable. So, to find the gradient of \(f(x, y)\), we need to compute the partial derivative of \(f\) with respect to \(x\) and the partial derivative of \(f\) with respect to \(y\). The gradient is denoted as \(abla f(x, y)\).

Step 2: Compute Partial Derivative with respect to x

First, we compute \(\frac{\partial f}{\partial x}\). The given function is \(f(x, y) = \sqrt{x^3 - 3xy}\). We apply the chain rule. Let \( u = x^3 - 3xy \), thus \( f(x, y) = u^{1/2} \). Therefore, \( \frac{\partial f}{\partial x} = \frac{1}{2} u^{-1/2} \left(3x^2 - 3y\right) = \frac{3x^2 - 3y}{2\sqrt{x^3 - 3xy}} \).

Step 3: Compute Partial Derivative with respect to y

Next, we compute \(\frac{\partial f}{\partial y}\). Again, using the same \(u\), the partial derivative is \( \frac{\partial f}{\partial y} = \frac{1}{2} u^{-1/2} \left(-3x\right) = \frac{-3x}{2\sqrt{x^3 - 3xy}} \).

Step 4: Combine Partial Derivatives into Gradient

Finally, we can write the gradient as a vector combining both partial derivatives: \(abla f(x, y) = \left( \frac{3x^2 - 3y}{2\sqrt{x^3 - 3xy}}, \frac{-3x}{2\sqrt{x^3 - 3xy}} \right)\).

Key Concepts

Gradient

In calculus, the gradient is more than just a measure of change; it's a way to understand how a function varies at any point in a multi-variable landscape.
When discussing a function of two or more variables, such as our example function \( f(x, y) \), the gradient provides a vector that points in the direction of the steepest increase of the function.
The components of the gradient vector are the partial derivatives with respect to each variable.
The gradient is symbolized as \( abla f(x, y) \).

• This vector gives us the direction in which the function \( f \) increases most rapidly.
• The magnitude of the gradient shows how steep the ascent is.

Thus, calculating the gradient involves determining partial derivatives, integrating various calculus concepts into one cohesive idea that points towards how a function behaves in different directions.

Partial Derivative

Partial derivatives are a cornerstone in multivariable calculus, allowing us to understand how a function changes with respect to one variable while keeping others fixed.
For the function \( f(x, y) \), partial derivatives are denoted by \( \frac{\partial f}{\partial x} \) and \( \frac{\partial f}{\partial y} \).

• \( \frac{\partial f}{\partial x} \) reflects how \( f \) changes as \( x \) varies, holding \( y \) constant.
• Conversely, \( \frac{\partial f}{\partial y} \) indicates how \( f \) changes with variations in \( y \), keeping \( x \) unchanged.

In our exercise, by computing these partial derivatives, we unravel the localized behavior of \( f(x, y) \) near any point \( (x, y) \).
Partial derivatives provide a detailed insight into individual variable impacts in functions of multiple variables, highlighting each variable's influence independently.

Chain Rule

The chain rule is an essential calculus tool to differentiate composite functions.
It provides a systematic method to break down derivatives into relatable parts, especially useful in complex expressions like \( f(x, y) = \sqrt{x^3 - 3xy} \).
To apply the chain rule, identify the inner and outer functions.

• Consider \( u = x^3 - 3xy \) as the inner function.
• The outer function becomes \( u^{1/2} \).

Taking the derivative of \( f \) becomes about following the order: compute the derivative of the outer function, then multiply by the derivative of the inner function.
This approach allows us to efficiently find each partial derivative by working through layers of complexity step-by-step.
Through the chain rule, we smoothly navigate multi-layered functions, simplifying the process into more manageable parts.