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Chapter 12: Problem 18

Find the first partial derivatives of the following functions. $$h(x, y)=\left(y^{2}+1\right) e^{x}$$

Short Answer

Expert verified
Question: Find the first partial derivatives of the function \(h(x, y) = (y^2 + 1)e^x\). Solution: The first partial derivatives of the given function are: $$\frac{\partial h}{\partial x} = (y^2 + 1)e^x$$ $$\frac{\partial h}{\partial y} = 2ye^x$$

Step by step solution

01

Identify the components of the function

First, identify the two functions that make up the given function. Let f(x) = e^x and g(y) = (y^2 + 1). The function h(x, y) can then be written as a product of f(x) and g(y): $$h(x, y) = f(x)g(y) = (y^2 + 1)e^x$$
02

Differentiate f(x) with respect to x

Next, find the partial derivative of f(x) with respect to x, denoted as \(\frac{\partial f}{\partial x}\): $$\frac{\partial f}{\partial x} = \frac{d}{dx} (e^x) = e^x$$
03

Differentiate g(y) with respect to y

Similarly, find the partial derivative of g(y) with respect to y, denoted as \(\frac{\partial g}{\partial y}\). Apply the power rule and sum rule of differentiation: $$\frac{\partial g}{\partial y} = \frac{d}{dy} (y^2 + 1) = 2y$$
04

Find the first partial derivative with respect to x

Now we'll find the first partial derivative of h(x, y) with respect to x, denoted as \(\frac{\partial h}{\partial x}\). Use the product rule of differentiation and the results from steps 2 and 3: $$\frac{\partial h}{\partial x} = (y^2 + 1)\cdot \frac{\partial f}{\partial x} + e^x\cdot \frac{\partial g}{\partial y}\big|_{\partial g/\partial y = 0}$$ $$\frac{\partial h}{\partial x} = (y^2 + 1)e^x$$
05

Find the first partial derivative with respect to y

Lastly, find the first partial derivative of h(x, y) with respect to y, denoted as \(\frac{\partial h}{\partial y}\). Again, use the product rule of differentiation and the results from steps 2 and 3: $$\frac{\partial h}{\partial y} = e^x \cdot \frac{\partial g}{\partial y} + (y^2 + 1)\cdot \frac{\partial f}{\partial x}\big|_{\partial f/\partial x = 0}$$ $$\frac{\partial h}{\partial y} = 2ye^x$$ The first partial derivatives of the given function are: $$\frac{\partial h}{\partial x} = (y^2 + 1)e^x$$ $$\frac{\partial h}{\partial y} = 2ye^x$$

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Key Concepts

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

Product Rule in Partial Derivatives
The product rule is a fundamental concept in calculus used to differentiate products of functions. When dealing with functions like \( h(x, y) = (y^2 + 1)e^x \), which is a product of \( f(x) = e^x \) and \( g(y) = y^2 + 1 \), the product rule helps find partial derivatives.

To apply this rule, think of \( h(x, y) \) as a product of two parts that are each differentiated with respect to a variable. For partial differentiation, here's how the rule works:
  • Differentiate one function while keeping the other constant.
  • Add the product of the derivatives.
The formula for the product rule in partial derivatives for \( rac{ ext{d}}{ ext{dx}}(uv) \) is:
  • \( u'v + uv' \)
where \( u \) and \( v \) are functions of \( x \). Apply this to \( u(x) = e^x \) and \( v(y) = y^2 + 1 \), making sure to follow the rule carefully for each variable.
Understanding Differentiation
Differentiation is the process of finding the derivative of a function, which represents how a function changes as its input changes. This is crucial in calculus for understanding rates of change and slopes of curves.

When differentiating, especially partial derivatives, remember these steps:
  • Identify the function components with respect to the chosen variable.
  • Apply differentiation rules like the power rule and sum rule.
For example, differentiating \( f(x) = e^x \) with respect to \( x \) gives the same function, \( e^x \). For \( g(y) = y^2 + 1 \), applying the power rule yields \( 2y \).

In partial differentiation, you handle each variable separately, treating the other variables as constants. This method simplifies finding derivatives, especially in functions of multiple variables.
The Role of Exponential Functions in Calculus
Exponential functions, like \( e^x \), play a key role in calculus due to their unique properties. Notably, the exponential function is its own derivative, making it especially interesting for differentiation.

Key properties to remember include:
  • The derivative of \( e^x \) is \( e^x \).
  • Exponential functions grow rapidly, showcasing constant relative growth.

In calculus exercises, exponential functions often appear alongside other functions, requiring careful application of differentiation rules. For instance, when differentiating \( (y^2 + 1)e^x \), remember that the exponential part \( e^x \) remains unchanged when differentiated with respect to \( x \). Understanding these properties simplifies finding derivatives and helps in solving complex problems involving exponential terms.

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