Question

Derivative of a multivariable composite function. For the function \(f(x, y(v))=x^{2} y+y^{3}\), where \(y=m v^{2}\), compute \(d f / d v\) around the point where \(m=1, v=2\), and \(x=3\).

Short answer

The derivative at the given point is 228.

Step-by-step solution

Identify the composite function components

The given function is: \[ f(x, y(v)) = x^2 y + y^3 \]and the inner function is: \[ y = m v^2 \].

Substitute the given values into the inner function

Given \( m = 1 \) and \( v = 2 \): \[ y = (1) (2)^2 = 4 \].

Substitute the given values into the outer function

Given \( x = 3 \): \[ f(3, 4) = (3)^2 (4) + (4)^3 = 9 \times 4 + 64 = 100 \].

Calculate the partial derivatives

Calculate the partial derivatives of \( f \): \[ \frac{\partial f}{\partial x} = 2xy \]\[ \frac{\partial f}{\partial y} = x^2 + 3y^2 \].

Evaluate the partial derivatives at the specific point

Evaluate the partial derivatives at \( x = 3 \), \( y = 4 \): \[ \frac{\partial f}{\partial x} = 2(3)(4) = 24 \]\[ \frac{\partial f}{\partial y} = (3)^2 + 3(4)^2 = 9 + 48 = 57 \].

Compute the derivative of the inner function with respect to \( v \)

The derivative of \( y \) with respect to \( v \) is: \[ \frac{dy}{dv} = \frac{d}{dv} (mv^2) = 2mv \].Given \( m = 1 \), evaluate at \( v = 2 \): \[ \frac{dy}{dv} = 2(1)(2) = 4 \].

Apply the chain rule

The chain rule for \( \frac{df}{dv} \) is: \[ \frac{df}{dv} = \frac{\partial f}{\partial x} \frac{dx}{dv} + \frac{\partial f}{\partial y} \frac{dy}{dv} \].Since \( x \) is constant with respect to \( v \), \( \frac{dx}{dv} = 0 \): \[ \frac{df}{dv} = \frac{57}{dv} \frac{4}{dv} = 57(4) = 228 \].

Key concepts

Partial Derivatives

Partial derivatives are an essential concept in multivariable calculus. They measure the rate at which a function changes as one of its variables changes, while keeping the other variables constant.
For a function of two variables, say, \(f(x, y)\), the partial derivatives with respect to \(x\) and \(y\) are denoted by \(\frac{\partial f}{\partial x}\) and \(\frac{\partial f}{\partial y}\), respectively.
To compute these, treat all other variables as constants and differentiate with respect to the chosen variable.

In our example, the partial derivatives of \(f(x, y) = x^2 y + y^3\) are:
\[ \frac{\partial f}{\partial x} = 2xy \]
\[ \frac{\partial f}{\partial y} = x^2 + 3y^2 \].
These tell us how \(f\) changes as \(x\) or \(y\) changes, respectively.

Chain Rule

The chain rule is a powerful tool for finding the derivative of a composite function.
When you have a function composed of other functions, like \(f(x, y(v))\), the chain rule helps compute the overall derivative by considering the derivatives of these inner functions.

The chain rule for one variable essentially states:
\[ \frac{df}{dx} = \frac{df}{dy} \cdot \frac{dy}{dx} \]
This means that the total derivative \(\frac{df}{dx}\) is the product of the derivative of \(f\) with respect to \(y\) and the derivative of \(y\) with respect to \(x\).

For multivariable functions, the chain rule extends to:
\[ \frac{df}{dv} = \frac{\partial f}{\partial x} \cdot \frac{dx}{dv} + \frac{\partial f}{\partial y} \cdot \frac{dy}{dv} \]
This formula helps us compute the derivative of \(f\) with respect to \(v\) by taking into account how both \(x\) and \(y\) change with \(v\).

Composite Functions

Composite functions are functions made up of other functions, where the output of one function is used as the input for another.

Consider our problem's function \(f(x, y(v)) = x^2 y + y^3\) where \(y = mv^2\).
Here, \(f\) depends on \(y\), which in turn depends on \(v\). This makes \(f\) a composite function.

Understanding composite functions is crucial when finding derivatives because changes in the outer function depend on changes in the inner function.
In simpler terms, knowing how one function feeds into another allows us to use tools like the chain rule to compute derivatives easily.

In the example:

• The inner function was \(y = mv^2\).
• We found \(y = 4\) by substituting \(m = 1\) and \(v = 2\).
• This value of \(y\) was then plugged into the outer function \(f(3, 4) = 100\).

This is a clear workflow of how composite functions operate.