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Magazine Chapter 14: Problem 42
Find all the second-order partial derivatives of the functions. \(f(x, y)=\sin x y\)
Short Answer
Expert verified
The second-order derivatives are: \(\frac{\partial^2 f}{\partial x^2} = -\sin(xy) y^2\), \(\frac{\partial^2 f}{\partial y^2} = -\sin(xy) x^2\), and \(\frac{\partial^2 f}{\partial y \partial x} = \frac{\partial^2 f}{\partial x \partial y} = -\sin(xy) xy + \cos(xy)\).
Step by step solution
01
Calculate First-Order Partial Derivative with respect to x
To find the partial derivative of \(f(x, y) = \sin(xy)\) with respect to \(x\), treat \(y\) as a constant. The derivative of \(\sin(u)\) where \(u = xy\) with respect to \(x\) is \(\cos(u) \cdot \frac{du}{dx}\). Here, \(\frac{du}{dx} = y\). Thus, \(\frac{\partial f}{\partial x} = \cos(xy) \cdot y\).
02
Calculate First-Order Partial Derivative with respect to y
Next, find the partial derivative of \(f(x, y) = \sin(xy)\) with respect to \(y\), treating \(x\) as a constant. Here, \(u = xy\) and \(\frac{du}{dy} = x\). Thus, \(\frac{\partial f}{\partial y} = \cos(xy) \cdot x\).
03
Calculate Second-Order Partial Derivative with respect to x twice
Now compute the second-order partial derivative with respect to \(x\) twice. Begin from \(\frac{\partial f}{\partial x} = \cos(xy) \cdot y\). Take the derivative with respect to \(x\) again: \[\frac{\partial^2 f}{\partial x^2} = \left( \frac{\partial}{\partial x}(\cos(xy) \cdot y) \right) = -\sin(xy) \cdot y^2\].
04
Calculate Mixed Second-Order Partial Derivative (first with respect to x, then y)
Compute the mixed second-order derivative, starting from \(\frac{\partial f}{\partial x} = \cos(xy) \cdot y\). Now differentiate with respect to \(y\): \[\frac{\partial^2 f}{\partial y \partial x} = \left( \frac{\partial}{\partial y}(\cos(xy) \cdot y) \right) = -\sin(xy) \cdot xy + \cos(xy)\].
05
Calculate Mixed Second-Order Partial Derivative (first with respect to y, then x)
For symmetry's sake, compute the mixed-order derivative beginning from \(\frac{\partial f}{\partial y} = \cos(xy) \cdot x\). Differentiate with respect to \(x\): \[\frac{\partial^2 f}{\partial x \partial y} = \left( \frac{\partial}{\partial x}(\cos(xy) \cdot x) \right) = -\sin(xy) \cdot xy + \cos(xy)\]. This result matches the mixed derivative calculated in Step 4, confirming the identity of mixed partial derivatives.
06
Calculate Second-Order Partial Derivative with respect to y twice
Lastly, compute the second-order partial derivative with respect to \(y\) twice. From \(\frac{\partial f}{\partial y} = \cos(xy) \cdot x\), differentiate with respect to \(y\) again: \[\frac{\partial^2 f}{\partial y^2} = \left( \frac{\partial}{\partial y}(\cos(xy) \cdot x) \right) = -\sin(xy) \cdot x^2\].
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Partial Derivatives
Partial derivatives are taken by focusing on one variable at a time while treating other variables as constants. For a function like \(f(x, y) = \sin(xy)\), to find the partial derivative with respect to \(x\), keep \(y\) constant. This means you differentiate just as if \(x\) is the only variable. Similarly, treat \(x\) as constant for the partial derivative with respect to \(y\).
- First-order partial derivatives give us the rate of change of the function with respect to one variable.
- Second-order partial derivatives involve differentiating the first-order derivatives again with respect to the same or another variable.
Mixed Derivatives
Mixed derivatives involve taking the derivative of a function with respect to more than one variable. This means applying the differentiation process in sequence for different variables, not once, but twice with adjustment.
For instance, for \(f(x, y) = \sin(xy)\), you first find the partial derivative with respect to \(x\), then take the derivative of that result with respect to \(y\), and vice versa.
For instance, for \(f(x, y) = \sin(xy)\), you first find the partial derivative with respect to \(x\), then take the derivative of that result with respect to \(y\), and vice versa.
- Mixed partial derivatives, such as \(\frac{\partial^2 f}{\partial x \partial y}\) and \(\frac{\partial^2 f}{\partial y \partial x}\), often end up being equal due to the Clairaut's theorem on equality of mixed partials. This theorem holds valid when the derivatives are continuous.
Trigonometric Functions
Trigonometric functions such as sine, cosine, and tangent are fundamental in calculus, especially when combined with functions of more than one variable. Here, the function \(f(x, y) = \sin(xy)\) is a beautiful example of how trigonometric functions integrate with multivariable calculus.
- The derivative of \(\sin(u)\) is \(\cos(u)\), which plays directly into how partial derivatives are calculated with functions involving sine and cosine.
- Understanding trigonometric identities can deeply aid in simplifying complex trigonometric expressions after differentiation.
Derivative Rules
In calculus, derivative rules are guidelines that help us find derivatives systematically. For functions like \(f(x, y) = \sin(xy)\) involving products and compositions, knowing these rules is very beneficial.
- The Chain Rule comes in handy when dealing with composite functions, as in \(\sin(xy)\). It states that the derivative of a function \(\sin(u)\) is \(\cos(u) \cdot \frac{du}{dx}\).
- The Product Rule is valuable when differentiating products of functions, like \(x \cdot \cos(xy)\).
