Question

If \(f(x, y, z)=\left(x^{3}+y^{2}+z\right)^{4}\), find each of the following: (a) \(f_{x}(x, y, z)\) (b) \(f_{y}(0,1,1)\) (c) \(f_{z z}(x, y, z)\)

Short answer

(a) 12x^2(x^3+y^2+z)^3; (b) 64; (c) 12(x^3+y^2+z)^2;

Step-by-step solution

Identify Part (a) Requirement

We need to find the partial derivative of the function \(f(x, y, z) = (x^3 + y^2 + z)^4\) with respect to \(x\). This involves treating \(y\) and \(z\) as constants and differentiating with respect to \(x\) only.

Differentiate for Part (a)

Use the chain rule to differentiate: let \(u = x^3 + y^2 + z\), so \(f(x, y, z) = u^4\). The derivative \(\frac{df}{dx} = 4u^3 \cdot \frac{du}{dx}\). Calculate \(\frac{du}{dx} = 3x^2\), then \(f_x = 4(x^3 + y^2 + z)^3 \cdot 3x^2\). Thus, \(f_x(x, y, z) = 12x^2(x^3 + y^2 + z)^3\).

Identify Part (b) Requirement

We need to find the partial derivative of the function with respect to \(y\) at the point (0, 1, 1). This means differentiating \(f(x, y, z)\) with respect to \(y\) and evaluating the derivative at the given point.

Differentiate and Evaluate for Part (b)

Differentiate using the chain rule: \(\frac{df}{dy} = 4(x^3 + y^2 + z)^3 \cdot \frac{du}{dy}\). Calculate \(\frac{du}{dy} = 2y\), which gives \(f_y = 8y(x^3 + y^2 + z)^3\). Evaluate at \((0, 1, 1)\): \(f_y(0, 1, 1) = 8(1)((0)^3 + 1^2 + 1)^3 = 8(1)(2)^3 = 64\).

Identify Part (c) Requirement

We need to find the second partial derivative of \(f(x, y, z)\) with respect to \(z\). This involves differentiating \(f_z\) with respect to \(z\) again.

Differentiate for Part (c)

First, compute \(f_z = 4(x^3 + y^2 + z)^3 \cdot 1 = 4(x^3 + y^2 + z)^3\). Now differentiate \(f_z\) with respect to \(z\): \(\frac{d^2f}{dz^2} = 12(x^3 + y^2 + z)^2 \cdot \frac{du}{dz} = 12(x^3 + y^2 + z)^2\) since \(\frac{du}{dz} = 1\). Therefore, \(f_{zz}(x, y, z) = 12(x^3 + y^2 + z)^2\).

Key concepts

Chain Rule

The chain rule is a fundamental technique in calculus that helps us differentiate composite functions. When dealing with multiple variables, it becomes vital in finding partial derivatives. The chain rule simplifies the differentiation process by breaking it down into easier steps. Consider a function like \[f(x, y, z) = (x^3 + y^2 + z)^4\]and assume we want to find its partial derivative with respect to a variable, say, \(x\). We treat the inside function \[u = x^3 + y^2 + z\]as a single entity or unit. The function can thus be rewritten in terms of \(u\) as \[f(u) = u^4\]. Applying the chain rule, if \(u\) is a function of \(x\), our derivative \(\frac{df}{dx}\) would be \[4u^3 \, \frac{du}{dx}\]. We then find \(\frac{du}{dx}\) by differentiating \(u\) with respect to \(x\), which gives \(3x^2\). Multiply them together to find \[f_x = 12x^2(x^3 + y^2 + z)^3\]. Notice how using the chain rule divides the process into smaller, more manageable pieces. Breaking it down helps isolate how each variable contributes to the derivative.

Multivariable Calculus

Multivariable calculus extends the basic principles of calculus into functions with more than one variable, like \(f(x, y, z)\). When dealing with such functions, we are frequently interested in understanding how changes in one variable affect the function while keeping the others constant. This is achieved through partial derivatives.Partial derivatives are the derivatives of functions with several variables. For instance, in the given function, we calculate partial derivatives like \(f_x, f_y,\) and \(f_z\) to find the rate of change of \(f\) with respect to \(x, y,\) and \(z\). Partial derivatives can be evaluated at specific points to get numerical values; for example, finding \(f_y\) at the point \((0, 1, 1)\) evaluates the influence of \(y\) at that exact location.In multivariable calculus, we use the chain rule to handle intricate expressions and variable dependencies. Multivariable calculus allows a nuanced exploration of systems encountered in fields ranging from physics to economics, treating each variable's influence either separately or collectively.

Second Partial Derivative

The second partial derivative refers to the derivative of a first partial derivative. In essence, it shows how the rate of change itself changes. It is beneficial in identifying the curvature and behavior of multivariable functions.For example, when we calculate \(f_{zz}\), we first find \(f_z = 4(x^3 + y^2 + z)^3\). Taking the derivative of \(f_z\)with respect to \(z\)gives us the second partial derivative: \[f_{zz} = 12(x^3 + y^2 + z)^2.\]Second partial derivatives can tell us about concavity or convexity at various points of a surface defined by the function. They are also instrumental in determining local maxima, minima, or saddle points in optimization problems, providing insight far beyond what first derivatives indicate.By calculating second partial derivatives, we delve deeper into the function's intricate details, exploring its nuanced geometric and physical interpretations.