Question

Find the gradient \(\nabla f\). $$ f(x, y)=\sin ^{3}\left(x^{2} y\right) $$

Short answer

The gradient is \( \nabla f = (3 \sin^2(x^2 y) \cos(x^2 y) \cdot 2x y, \, 3 \sin^2(x^2 y) \cos(x^2 y) \cdot x^2) \).

Step-by-step solution

Understand the Function

We have a function \( f(x, y) = \sin^3(x^2 y) \). To find the gradient, we need to calculate the partial derivatives with respect to \(x\) and \(y\).

Apply the Chain Rule

The function \( f(x, y) = \sin^3(x^2 y) \) is composed of an outer function, \( ( ext{something})^3 \), and an inner function \( \sin(x^2 y) \). Use the chain rule to differentiate: if \( u = x^2 y \) and \( v = \sin(u) \), then \( f = v^3 \).

Partial Derivative with Respect to \(x\)

Differentiate \( f(x, y) = (\sin(x^2 y))^3 \) with respect to \(x\):\[ \text{First differentiate the outer part: } 3 \sin^2(x^2 y) \times \cos(x^2 y). \]\[ \text{Then differentiate the inner part } x^2 y \, ext{ with respect to } x: 2x y.\]Thus, the partial derivative with respect to \(x\) is:\[ \frac{\partial f}{\partial x} = 3 \sin^2(x^2 y) \cdot \cos(x^2 y) \cdot (2x y). \]

Partial Derivative with Respect to \(y\)

Differentiate \( f(x, y) = (\sin(x^2 y))^3 \) with respect to \(y\):\[ \text{First differentiate the outer part: } 3 \sin^2(x^2 y) \times \cos(x^2 y). \]\[ \text{Then differentiate the inner part } x^2 y \, ext{ with respect to } y: x^2. \]Thus, the partial derivative with respect to \(y\) is:\[ \frac{\partial f}{\partial y} = 3 \sin^2(x^2 y) \cdot \cos(x^2 y) \cdot x^2. \]

Write the Gradient

The gradient \( abla f \) consists of the partial derivatives \( \frac{\partial f}{\partial x} \) and \( \frac{\partial f}{\partial y} \). Therefore, we have:\[ abla f = \left( 3 \sin^2(x^2 y) \cos(x^2 y) \cdot 2x y, \, 3 \sin^2(x^2 y) \cos(x^2 y) \cdot x^2 \right). \]

Key concepts

Partial Derivatives

In calculus, partial derivatives are a way to measure how a multivariable function changes as we change one of the variables, keeping the others constant. Think of it as peeking into the "slope" of a function at a particular point along an axis.
For the function \( f(x, y) = \sin^3(x^2 y) \), we take partial derivatives to find how \( f \) changes when we slightly change \( x \) or \( y \).
To do this, you treat one variable as a constant, while differentiating with respect to the other. Here's a step-by-step:

• For \( \frac{\partial f}{\partial x} \), keep \( y \) constant and differentiate \( f \) with respect to \( x \).
• For \( \frac{\partial f}{\partial y} \), keep \( x \) constant and differentiate \( f \) with respect to \( y \).

These derivatives give you the components of the gradient, which shows the direction of the steepest ascent in a multivariable landscape.

Chain Rule in Calculus

The chain rule is a fundamental concept in calculus, used to differentiate compositions of functions. It's like peeling layers of an onion, delving into each layer independently.
When dealing with a compound function like \( f(x, y) = \sin^3(x^2 y) \), you can visualize it as an outer layer (\( (\text{something})^3 \)) and an inner layer (\( \sin(x^2 y) \)).
Here's how you apply the chain rule:

• Differentiate the outer function, treating the inside as a single entity.
• Multiply by the derivative of the inner function (each variable separately, if needed).

For example, differentiating \( f(x, y) \) with respect to \( x \) involves:
1. Differentiating \( 3\sin^2(x^2 y) \) as the outer function.
2. Multiplying by \( \cos(x^2 y) \) from differentiating \( \sin(x^2 y) \).
3. Multiplying by \( 2x y \), the derivative of \( x^2 y \) with respect to \( x \).
This approach handles the intricacies of multivariable differentiation elegantly.

Multivariable Calculus

Multivariable calculus extends regular calculus into spaces with more than one dimension, handling functions with multiple variables.
Concepts like the gradient rely heavily on understanding these multiple dimensions.
The gradient, \( abla f \), is a vector that points in the direction of the greatest rate of increase of the function. It consists of partial derivatives of each variable:

• The \( x \)-component is \( \frac{\partial f}{\partial x} \).
• The \( y \)-component is \( \frac{\partial f}{\partial y} \).

For the function \( f(x, y) = \sin^3(x^2 y) \), the gradient describes how steep the slope is and which way is uphill when standing at a point \((x, y)\).
This provides critical insights, helping resolve various practical problems from logistics optimization to physics simulations. Multivariable calculus bridges gaps in understanding complex systems by simplifying them into manageable pieces.