Problem 12
Question
Find the first partial derivatives with respect to \(x\) and with respect to \(y .\) $$ g(x, y)=e^{x / y} $$
Step-by-Step Solution
Verified Answer
\[\frac{\partial g}{\partial x} =h e^{x/y} / y, \quad \frac{\partial g}{\partial y} = -x \cdot e^{x / y} / y^{2}\]
1Step 1: Compute the partial derivative with respect to x
Applying the chain rule gives: \[\frac{\partial g}{\partial x} = e^{x/y} \cdot \frac{\partial (x/y)}{\partial x} = e^{x/y} \cdot (1/y)\]
2Step 2: Compute the partial derivative with respect to y
Again, apply the chain rule, bearing in mind that this time y is the variable, and \(x/y\) will have to be treated as a function of y: \[\frac{\partial g}{\partial y} = e^{x/y} \cdot \frac{\partial (x/y)}{\partial y}\] However, now the reciprocal rule must also be used: \[\frac{\partial (x/y)}{\partial y} = x \cdot (-1/y^2)\] thus: \[\frac{\partial g}{\partial y} = e^{x/y} \cdot (-x/y^2) = -x \cdot e^{x / y} / y^{2}\]
Key Concepts
Chain RuleReciprocal RuleMultivariable CalculusExponential Functions
Chain Rule
The chain rule is a fundamental formula in calculus used when differentiating composite functions. It essentially tells us that if we have a function of a function, the derivative of the overall expression is the derivative of the outer function, multiplied by the derivative of the inner function.
When dealing with multivariable calculus, understanding how the chain rule works is crucial. For instance, when trying to find the derivative of a function like \( g(x, y) = e^{x / y} \) with respect to \( x \), we treat \( x/y \) as a separate function inside the exponential function. Hence, according to the chain rule, the partial derivative with respect to \( x \) is the exponent times the derivative of \( x/y \) with respect to \( x \), leading to the simplified form of \( \frac{\text{d}}{\text{d}x}(e^{x/y}) = e^{x/y} \cdot (1/y) \).
When dealing with multivariable calculus, understanding how the chain rule works is crucial. For instance, when trying to find the derivative of a function like \( g(x, y) = e^{x / y} \) with respect to \( x \), we treat \( x/y \) as a separate function inside the exponential function. Hence, according to the chain rule, the partial derivative with respect to \( x \) is the exponent times the derivative of \( x/y \) with respect to \( x \), leading to the simplified form of \( \frac{\text{d}}{\text{d}x}(e^{x/y}) = e^{x/y} \cdot (1/y) \).
Reciprocal Rule
The reciprocal rule is a powerful tool in calculus that makes it easier to differentiate functions that are in the form of a reciprocal, such as \( 1/y \) or \( x/y \).
The rule states that if you want to differentiate \( 1/f(x) \), where \( f(x) \) is a differentiable function, you can use the formula \( -f'(x) / [f(x)]^2 \). In our exercise, to find the partial derivative of \( x/y \) with respect to \( y \), we see that it can be written as \( x \cdot (1/y) \). Applying the reciprocal rule, we derive that \( \frac{\text{d}}{\text{d}y}(x/y) = -x/y^2 \), which simplifies the differentiation process considerably.
The rule states that if you want to differentiate \( 1/f(x) \), where \( f(x) \) is a differentiable function, you can use the formula \( -f'(x) / [f(x)]^2 \). In our exercise, to find the partial derivative of \( x/y \) with respect to \( y \), we see that it can be written as \( x \cdot (1/y) \). Applying the reciprocal rule, we derive that \( \frac{\text{d}}{\text{d}y}(x/y) = -x/y^2 \), which simplifies the differentiation process considerably.
Multivariable Calculus
Multivariable calculus expands on single-variable calculus by considering functions of two or more variables. In this broader context, partial derivatives become an essential tool, as they allow us to measure how a function changes as each variable is varied while keeping the others fixed.
In the given exercise, \( g(x, y) = e^{x / y} \) is a function of two variables, \( x \) and \( y \). Here, we're interested in how \( g \) changes with respect to each variable independently. To do this, we compute the first partial derivatives with respect to \( x \) and \( y \), demonstrating how multivariable functions often require the application of rules from single-variable calculus in a more intricate setting.
In the given exercise, \( g(x, y) = e^{x / y} \) is a function of two variables, \( x \) and \( y \). Here, we're interested in how \( g \) changes with respect to each variable independently. To do this, we compute the first partial derivatives with respect to \( x \) and \( y \), demonstrating how multivariable functions often require the application of rules from single-variable calculus in a more intricate setting.
Exponential Functions
Exponential functions are pervasive in various fields such as mathematics, physics, and economics. These are functions of the form \( f(x) = a^x \), where \( a \) is a constant, and they are characterized by their rapid growth.
The natural exponential function, where \( a \) is the mathematical constant \( e \), is written as \( e^x \), and it has the unique property that the rate of change is proportional to the function's value at any point. This is why the derivative of \( e^x \) is simply \( e^x \). In our exercise, \( g(x, y) = e^{x / y} \) is an exponential function where the exponent is a quotient, which results in a more complex differentiation that necessitates the use of the chain and reciprocal rules.
The natural exponential function, where \( a \) is the mathematical constant \( e \), is written as \( e^x \), and it has the unique property that the rate of change is proportional to the function's value at any point. This is why the derivative of \( e^x \) is simply \( e^x \). In our exercise, \( g(x, y) = e^{x / y} \) is an exponential function where the exponent is a quotient, which results in a more complex differentiation that necessitates the use of the chain and reciprocal rules.
Other exercises in this chapter
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