Question

Find all first partial derivatives of each function. \(f(x, y)=\left(3 x^{2}+y^{2}\right)^{-1 / 3}\)

Short answer

The partial derivatives are: \( \frac{\partial f}{\partial x} = -2x(3x^2 + y^2)^{-4/3} \) and \( \frac{\partial f}{\partial y} = -\frac{2y}{3}(3x^2 + y^2)^{-4/3} \).

Step-by-step solution

Determine the Partial Derivative with Respect to x

To find the partial derivative of the function \( f(x, y) = (3x^2 + y^2)^{-1/3} \) with respect to \( x \), first recognize it as a function of the form \( u^{n} \) where \( u = 3x^2 + y^2 \) and \( n = -1/3 \). Use the chain rule for differentiation: \( \frac{\partial f}{\partial x} = \frac{d}{du}[u^n] \cdot \frac{\partial u}{\partial x} \).\Calculate \( \frac{d}{du}[u^{-1/3}] = -\frac{1}{3}u^{-4/3} \).\Compute \( \frac{\partial u}{\partial x} = 6x \).\Thus, \( \frac{\partial f}{\partial x} = -\frac{1}{3}(3x^2 + y^2)^{-4/3} \cdot 6x \). After simplification, this becomes \( -2x(3x^2 + y^2)^{-4/3} \).

Determine the Partial Derivative with Respect to y

Next, find the partial derivative of the function \( f(x, y) = (3x^2 + y^2)^{-1/3} \) with respect to \( y \). Again, use the chain rule: \( \frac{\partial f}{\partial y} = \frac{d}{du}[u^n] \cdot \frac{\partial u}{\partial y} \), where \( u = 3x^2 + y^2 \) and \( n = -1/3 \).\Calculate \( \frac{d}{du}[u^{-1/3}] = -\frac{1}{3}u^{-4/3} \) (same as before).\Compute \( \frac{\partial u}{\partial y} = 2y \).\Thus, \( \frac{\partial f}{\partial y} = -\frac{1}{3}(3x^2 + y^2)^{-4/3} \cdot 2y \). After simplification, this becomes \( -\frac{2y}{3}(3x^2 + y^2)^{-4/3} \).

Key concepts

Chain Rule

The chain rule is a fundamental tool in calculus for differentiating compositions of functions. When dealing with multivariable functions like \( f(x, y) = (3x^2 + y^2)^{-1/3} \), the chain rule helps us find partial derivatives by focusing on how the derivative of one variable travels through the function via inner and outer functions.

For

• the partial derivative of a function \( f \) with respect to a variable like \( x \), the chain rule involves first differentiating the outer function \( u^n \), followed by multiplying it with the derivative of the inner function \( u \) with respect to \( x \), where \( u = 3x^2 + y^2 \).
• Differentiating steps lead to expressions such as \( \frac{d}{du}[u^{-1/3}] = -\frac{1}{3}u^{-4/3} \), and \( \frac{\partial u}{\partial x} = 6x \).

Thus, the chain rule in this context stitches together these derivatives to find the overall rate of change of \( f \) with respect to \( x \), given by \( -2x(3x^2 + y^2)^{-4/3} \).

This rule reverses with \( y \), where changing the inner derivative to \( 2y \) affects changes with respect to \( y \). Understanding how to interlink these derivatives is what makes the chain rule indispensable for complex equations in calculus.

Multivariable Calculus

Multivariable calculus extends single-variable calculus into functions with multiple variables. For instance, the function \( f(x, y) = (3x^2 + y^2)^{-1/3} \) involves two independent variables, \( x \) and \( y \). Each of these affects the dependent variable, \( f(x, y) \), separately and together.

Understanding multivariable calculus is crucial when evaluating functions that model more complex real-world phenomena. In practice, such functions appear in scenarios ranging from physics to economics.

A key learning point in multivariable calculus is how to take partial derivatives. It becomes essential to analyze the change of \( f \) by isolating changes in one variable at a time, explaining rates of change both individually and collectively. The method allows for precise measurements of sensitivity or responsiveness of one variable's change over another specific variable, keeping all other variables constant.

This approach provides insight into how a system responds to separate tweaks. For \( f(x, y) = (3x^2 + y^2)^{-1/3} \), each partial derivative tells how \( f \) changes if only \( x \) or \( y \) varies.

Differentiation

Differentiation is the process of finding a derivative, which measures how a function changes as its input changes. In the context of our multivariable function \( f(x, y) = (3x^2 + y^2)^{-1/3} \), differentiation helps us understand how this "change" can be expressed mathematically with respect to each variable.

In single-variable calculus, differentiation might be straightforward, often involving basic rules like the power rule or product rule. However, multivariable differentiation introduces complexity since a function might change differently along multiple variables.

In the given problem, we're using differentiation to calculate partial derivatives which tell us the rate of change of \( f \) with respect to just one of the variables at a time.

• For \( \frac{\partial f}{\partial x} \), we fix \( y \) and consider only how changes in \( x \) affect \( f \).
• For \( \frac{\partial f}{\partial y} \), \( x \) is fixed and we're solely looking at alterations in \( y \) affecting \( f \).

Differentiation in multivariable calculus elucidates these intricate interactions between variables, making it a cornerstone of understanding how variable interrelations shape outputs in complex systems.