Question

If \(u=x^{2}\left(x^{2}+y^{2}\right)\), find \(\partial u / \partial x, \partial u / \partial y\).

Short answer

W.r.t. x: \(4x^3 + 2x y^2\), W.r.t. y: \(2x^2 y\)

Step-by-step solution

- Identify the function to differentiate

The given function is: \[ u = x^2 (x^2 + y^2) \] We need to find the partial derivatives \( \frac{\partial u}{\partial x} \) and \( \frac{\partial u}{\partial y} \).

- Expand the function

First, expand \( u \) to simplify the differentiation process: \[ u = x^2 (x^2 + y^2) = x^2 \cdot x^2 + x^2 \cdot y^2 = x^4 + x^2 y^2 \]

- Differentiate with respect to \( x \)

Apply the power rule to each term in the expanded function: \[ \frac{\partial u}{\partial x} = \frac{\partial}{\partial x} (x^4 + x^2 y^2) = 4x^3 + 2x y^2 \]

- Differentiate with respect to \( y \)

Again, apply the power rule but this time with respect to \( y \): \[ \frac{\partial u}{\partial y} = \frac{\partial}{\partial y} (x^4 + x^2 y^2) = 0 + 2x^2 y = 2x^2 y \]

Key concepts

power rule

The power rule is crucial for finding derivatives of functions involving exponents. The rule states that if you have a function of the form \( f(x) = x^n \), then the derivative of this function with respect to x is \( f'(x) = nx^{n-1} \).
In our exercise, we used the power rule in step 3 when differentiating terms like \( x^4 \) and \( x^2 y^2 \) with respect to \( x \).
Here are the steps:

• For \( x^4 \), apply the power rule to get \( 4x^3 \).
• For \( x^2 y^2 \), treat \( y^2 \) as a constant while differentiating with respect to \( x \). The result is \( 2x y^2 \).

By applying the power rule, the differentiation becomes straightforward and manageable.

differentiation

Differentiation is a fundamental tool in calculus used to find how a function changes as its input changes. When dealing with single-variable functions, differentiation measures the rate of change with respect to one variable.
In our exercise, we applied differentiation to a multivariable function. Let's break it down:

• First, simplify the function for clearer differentiation.
• Then, apply differentiation rules to each term.

For partial derivatives, we treat other variables as constants while differentiating with respect to our variable of interest. For instance, \( y \) was considered a constant while differentiating with respect to \( x \). This method allowed us to find how the function \( u \) changes as \( x \) or \( y \) changes independently.

multivariable calculus

Multivariable calculus extends the concepts of differentiation and integration to functions with more than one variable. It involves tools like partial derivatives which measure how a multivariable function changes as each variable changes while keeping the others constant.
For the given function \( u = x^2 (x^2 + y^2) \), we found partial derivatives with respect to \( x \) and \( y \):
\( u = x^4 + x^2 y^2 \)
1. \( \frac{\text{partial } u}{\text{partial } x} = 4x^3 + 2x y^2 \)
2. \( \frac{\text{partial } u}{\text{partial } y} = 2x^2 y \)
These results show how \( u \) changes when we tweak \( x \) or \( y \) independently, shedding light on the behavior of complex systems described by multivariable functions.