Question

Find the partial derivatives with respect to \(x, y,\) and \(z\) of the following functions: (a) \(f(x, y, z)=\) \(a x^{2}+b x y+c y^{2},(\mathbf{b}) g(x, y, z)=\sin \left(a x y z^{2}\right),(\mathbf{c}) h(x, y, z)=a e^{x y / z^{2}},\) where \(a, b,\) and \(c\) are constants. Remember that to evaluate \(\partial f / \partial x\) you differentiate with respect to \(x\) treating \(y\) and \(z\) as constants.

Short answer

For function (a), partial derivatives: \( \frac{\partial f}{\partial x} = 2ax + by \), \( \frac{\partial f}{\partial y} = bx + 2cy \), \( \frac{\partial f}{\partial z} = 0 \). For function (b): \( \frac{\partial g}{\partial x} = ayz^2 \cos(ax yz^2) \), \( \frac{\partial g}{\partial y} = axz^2 \cos(ax yz^2) \), \( \frac{\partial g}{\partial z} = 2axyz \cos(ax yz^2) \). For function (c): \( \frac{\partial h}{\partial x} = a\frac{y}{z^2} e^{xy/z^2} \), \( \frac{\partial h}{\partial y} = a\frac{x}{z^2} e^{xy/z^2} \), \( \frac{\partial h}{\partial z} = -a\frac{2xy}{z^3} e^{xy/z^2} \).

Step-by-step solution

Find \( \frac{\partial f}{\partial x} \)

To find the partial derivative of the function \(f(x, y, z) = ax^2 + bxy + cy^2 \) with respect to \(x\):1. Treat \(y\) and \(z\) as constants.2. The derivative of \(ax^2\) with respect to \(x\) is \(2ax\).3. The derivative of \(bxy\) with respect to \(x\) is \(by\).4. The term \(cy^2\) is constant with respect to \(x\), so its derivative is \(0\).Thus, \( \frac{\partial f}{\partial x} = 2ax + by \).

Find \( \frac{\partial f}{\partial y} \)

To find the partial derivative of \(f(x, y, z) = ax^2 + bxy + cy^2 \) with respect to \(y\):1. Treat \(x\) and \(z\) as constants.2. The derivative of \(ax^2\) with respect to \(y\) is \(0\).3. The derivative of \(bxy\) with respect to \(y\) is \(bx\).4. The derivative of \(cy^2\) with respect to \(y\) is \(2cy\).Thus, \( \frac{\partial f}{\partial y} = bx + 2cy \).

Find \( \frac{\partial f}{\partial z} \)

To find the partial derivative of \(f(x, y, z) = ax^2 + bxy + cy^2 \) with respect to \(z\):1. Treat \(x\) and \(y\) as constants.2. As there are no terms in the function involving \(z\), all derivatives with respect to \(z\) are \(0\).Thus, \( \frac{\partial f}{\partial z} = 0 \).

Find \( \frac{\partial g}{\partial x} \)

To find the partial derivative of \(g(x, y, z) = \sin(ax yz^2)\) with respect to \(x\):1. Treat \(y\) and \(z\) as constants.2. Using the chain rule, derivative of \(\sin(u)\) is \(\cos(u)\) multiplied by the derivative of \(u\), for \(u = ax yz^2\).3. The derivative of \(u = ax yz^2\) with respect to \(x\) is \(ayz^2\).Thus, \( \frac{\partial g}{\partial x} = \cos(ax yz^2) \cdot ayz^2 \).

Find \( \frac{\partial g}{\partial y} \)

To find the partial derivative of \(g(x, y, z) = \sin(ax yz^2)\) with respect to \(y\):1. Treat \(x\) and \(z\) as constants.2. Use the chain rule: derivative of \(\sin(u)\) is \(\cos(u)\) multiplied by the derivative of \(u\), for \(u = ax yz^2\).3. The derivative of \(u = ax yz^2\) with respect to \(y\) is \(axz^2\).Thus, \( \frac{\partial g}{\partial y} = \cos(ax yz^2) \cdot axz^2 \).

Find \( \frac{\partial g}{\partial z} \)

To find the partial derivative of \(g(x, y, z) = \sin(ax yz^2)\) with respect to \(z\):1. Treat \(x\) and \(y\) as constants.2. Use the chain rule: derivative of \(\sin(u)\) is \(\cos(u)\) multiplied by the derivative of \(u\), for \(u = ax yz^2\).3. The derivative of \(u = ax yz^2\) with respect to \(z\) is \(2ax yz\).Thus, \( \frac{\partial g}{\partial z} = \cos(ax yz^2) \cdot 2ax yz \).

Find \( \frac{\partial h}{\partial x} \)

To find the partial derivative of \(h(x, y, z) = ae^{xy/z^2}\) with respect to \(x\):1. Treat \(y\) and \(z\) as constants.2. Using the chain rule: The derivative of \(e^u\) is \(e^u\) times the derivative of \(u\), for \(u = xy/z^2\).3. The derivative of \(u = xy/z^2\) with respect to \(x\) is \(y/z^2\).Thus, \( \frac{\partial h}{\partial x} = ae^{xy/z^2} \cdot \frac{y}{z^2} \).

Find \( \frac{\partial h}{\partial y} \)

To find the partial derivative of \(h(x, y, z) = ae^{xy/z^2}\) with respect to \(y\):1. Treat \(x\) and \(z\) as constants.2. Using the chain rule: The derivative of \(e^u\) is \(e^u\) times the derivative of \(u\), for \(u = xy/z^2\).3. The derivative of \(u = xy/z^2\) with respect to \(y\) is \(x/z^2\).Thus, \( \frac{\partial h}{\partial y} = ae^{xy/z^2} \cdot \frac{x}{z^2} \).

Find \( \frac{\partial h}{\partial z} \)

To find the partial derivative of \(h(x, y, z) = ae^{xy/z^2}\) with respect to \(z\):1. Treat \(x\) and \(y\) as constants.2. Using the chain rule: The derivative of \(e^u\) is \(e^u\) times the derivative of \(u\), for \(u = xy/z^2\).3. The derivative of \(u = xy/z^2\) with respect to \(z\) is \(-2xy/z^3\).Thus, \( \frac{\partial h}{\partial z} = ae^{xy/z^2} \cdot \left(-\frac{2xy}{z^3}\right) \).

Key concepts

Understanding the Chain Rule in Multivariable Calculus

When working with multivariable functions, the chain rule becomes an invaluable tool. It helps us differentiate functions composed of multiple sub-functions. Imagine you have a function like \(g(x, y, z) = \sin(ax yz^2)\). Here, the dependent variable depends on two or more variables. The chain rule can simplify the differentiation process by breaking the function into parts.
For a function \(f(u)\), where \(u\) is another function like \(u = ax yz^2\), the chain rule states that the derivative of \(f\) is the derivative of \(f\) with respect to \(u\) multiplied by the derivative of \(u\) with respect to the variable of interest.
In practice:

• Differentiate the outer function. For \(\sin(u)\), it becomes \(\cos(u)\).
• Next, find the derivative of the inner function \(u\) as it relates to your variable of interest, like \(x\), \(y\), or \(z\).
• Multiply these derivatives to get the final result.

The power of the chain rule lies in unraveling complex expressions into manageable components.

Exploring Multivariable Calculus

Multivariable calculus extends single-variable calculus to functions of two or more variables. In contrast to single-variable calculus, where you only deal with one independent variable, multivariable calculus involves several independent variables.
This allows for a more sophisticated description of the world, as most systems and processes are inherently multivariable.
With functions like \(f(x, y, z) = ax^2 + bxy + cy^2\), you can calculate how one variable uniquely influences the overall function while the others remain constant. This is done using partial derivatives, a central concept in multivariable calculus.
Applications of multivariable calculus are vast and include physics, engineering, economics, and any field where change occurs over more than one dimension. Understanding this helps model and predict complex systems by exploring interactions between variables.

Differentiation with Respect to a Variable

Differentiation with respect to a variable involves taking the derivative of a function concerning one variable, while treating others as constants. For example, in a function with three variables, like \(h(x, y, z) = ae^{xy/z^2}\), partial differentiation allows you to see how changing one variable will affect the function.
Here’s how it works:

• Choose the variable you want to differentiate with respect to.
• Treat the other variables as constants.
• Apply standard differentiation rules, like the power rule, product rule, or chain rule, as if the function only depended on this one variable.

This method simplifies the calculation and reveals the rate of change with respect to one variable without considering others. It's a powerful technique used for functions where multiple factors are at play, offering a deeper insight into how variables interact.