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Find the first partial derivatives of the function. $$ f(x, y)=x^{y} $$

Short Answer

Expert verified
The partial derivatives are: \(\frac{\partial f}{\partial x} = y \cdot x^{y-1}\) and \(\frac{\partial f}{\partial y} = x^y \ln(x)\).

Step by step solution

01

Understanding Partial Derivative with Respect to x

We want to find the partial derivative of the function with respect to \(x\). When taking the partial derivative with respect to \(x\), treat \(y\) as a constant.
02

Apply the Natural Log to Simplify

To differentiate the function \(f(x,y) = x^y\), it helps to use the natural logarithm. Consider \(z = x^y\), where \(z = e^{y \ln(x)}\). This transforms the expression to one easier to differentiate using chain and product rules.
03

Differentiate with Respect to x

Differentiate \(z = e^{y \ln(x)}\) with respect to \(x\). Use the chain rule, where \(\frac{dz}{dx} = e^{y \ln(x)} \cdot \frac{d}{dx}(y \ln(x)) = x^{y} \cdot \frac{y}{x}\). Simplifying gives us: \[\frac{\partial f}{\partial x} = y \cdot x^{y-1}\].
04

Understanding Partial Derivative with Respect to y

Next, we find the partial derivative with respect to \(y\). Here, treat \(x\) as a constant.
05

Differentiate with Respect to y

For \(f(x,y) = x^y\), differentiate with respect to \(y\). Use the rule for differentiating an exponential function: \(\frac{d}{dy}(x^y) = x^y \cdot \ln(x)\). Thus, we have:\[\frac{\partial f}{\partial y} = x^y \ln(x)\].

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Natural Logarithm Differentiation
The natural logarithm, often denoted as \( \ln(x) \), is an essential tool in calculus. When we deal with complex expressions, taking the natural logarithm can simplify differentiation. The natural logarithm transforms multiplication into addition, which is usually easier to handle when differentiating.

For instance, consider the function \( x^y \). To differentiate using logarithmic differentiation, we set \( z = x^y = e^{y \ln(x)} \). Taking the logarithm of both sides reduces the problem into simpler components. This process highlights the inner workings of the function and allows us to apply other differentiation rules effectively.

Remember, using natural logs is handy when dealing with exponential forms, making the subsequent use of the chain and product rules straightforward.
Chain Rule
The chain rule is a fundamental differentiation tool used when dealing with composite functions. This rule helps us find the derivative of a function nested within another function. For a function \( f(g(x)) \), the chain rule states that the derivative is \( f'(g(x)) \cdot g'(x) \).

In our example \( z = e^{y \ln(x)} \), where \( y \ln(x) \) is the inner function, and \( e^u \) is the outer function, the chain rule plays a crucial role. We differentiate the outer function with respect to its argument and then multiply by the derivative of the inner function.

Thus, the derivative becomes \( \frac{dz}{dx} = e^{y \ln(x)} \cdot \frac{d}{dx}(y \ln(x)) \). This rule ensures that we handle the layered structure of functions correctly, making it central to calculus.
Exponential Function Differentiation
The process of differentiation for exponential functions requires specific rules due to their unique properties. When differentiating exponential functions like \( x^y \), especially with variables involved in both the exponent and the base, it's vital to apply these rules precisely.

For the function \( x^y \), when differentiated with respect to \( y \), we consider \( x^y = e^{y \ln(x)} \). The exponential function differentiation states that \( \frac{d}{du}e^u = e^u \). Thus, differentiating with respect to \( y \) becomes straightforward due to the chain rule application.

In practice, this leads to the result \( x^y \cdot \ln(x) \) when differentiating \( x^y \) concerning \( y \). Understanding this method enables handling varied forms of exponential functions in calculus problems.
Product Rule
The product rule is a valuable method used to differentiate expressions where two or more functions are multiplied together. According to this rule, if we have two functions, \( u(x) \) and \( v(x) \), their derivative is given by \( u'(x)v(x) + u(x)v'(x) \).

When applying the product rule, it's essential first to identify each separate function and then find their respective derivatives.

In our case, you may initially think of the expression \( x^y \) in its logarithmic form, attempting to view it as a product. However, further simplification and differentiation generally involve other rules like the chain and exponential rules in more depth.

Though the direct role of the product rule may seem less apparent in some contexts, together with other rules, it forms the backbone of many differentiation strategies.

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Most popular questions from this chapter

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