Problem 23 - Calculus Early Transcendentals

Question

Find the first partial derivatives of the following functions. $$f(x, y)=x^{2 y}$$

Step-by-Step Solution

Answer

Question: Find the partial derivatives of the following function: \(f(x, y) = x^{2y}\) Solution: The function \(f(x, y) = x^{2y}\) has first partial derivatives with respect to \(x\) and \(y\) as follows: The partial derivative respect to \(x\) is: $$\frac{\partial f}{\partial x} = 2yx^{(2y-1)}$$ The partial derivative respect to \(y\) is: $$\frac{\partial f}{\partial y} = 2\ln x\cdot x^{2y}$$

Step 1: Partial Derivative with respect to x

To find the partial derivative of \(f(x, y) = x^{2y}\) with respect to \(x\), we will use the chain rule. Recall that the chain rule for a function of the form \(f(x) = u(v(x))\) is given by \(f'(x) = u'(v(x)) * v'(x)\). In our case, let \(u(x) = x^{2y}\) and \(v(x) = 2y\). We have: $$\frac{\partial f}{\partial x} = \frac{\partial (x^{2y})}{\partial x}$$ Using the chain rule, we get: $$\frac{\partial f}{\partial x} = (2y) x^{(2y-1)}$$ So, the partial derivative of \(f(x, y)\) with respect to \(x\) is: $$\frac{\partial f}{\partial x} = 2yx^{(2y-1)}$$

Step 2: Partial Derivative with respect to y

To find the partial derivative of \(f(x, y) = x^{2y}\) with respect to \(y\), we will first rewrite the function as an exponential function with base \(e\) using the identity \(a^b = e^{(\ln a)b}\): $$f(x, y) = x^{2y} = e^{(\ln x)(2y)}$$ Now, find the partial derivative with respect to \(y\), using the chain rule: $$\frac{\partial f}{\partial y} = \frac{\partial (e^{(\ln x)(2y)})}{\partial y}$$ The derivative of \(e^u\) with respect to \(u\) is \(e^u\). Using the chain rule, we have: $$\frac{\partial f}{\partial y} = e^{(\ln x)(2y)}\cdot \frac{\partial((\ln x)(2y))}{\partial y}$$ Now differentiate \((\ln x)(2y)\) with respect to \(y\): $$\frac{\partial((\ln x)(2y))}{\partial y} = 2\ln x$$ So, the partial derivative of \(f(x, y)\) with respect to \(y\) is: $$\frac{\partial f}{\partial y} = 2\ln x\cdot e^{(\ln x)(2y)} = 2\ln x\cdot x^{2y}$$ Now we have both first partial derivatives: $$\frac{\partial f}{\partial x} = 2yx^{(2y-1)}$$ $$\frac{\partial f}{\partial y} = 2\ln x\cdot x^{2y}$$

Key Concepts

Chain Rule

The chain rule is a fundamental tool in calculus that helps us differentiate composite functions. Simply put, if you have a function nested inside another function, the chain rule is how you find its derivative. For example, if you have a function defined as

• \( f(x) = u(v(x)) \)

To differentiate it, we find the derivative of the outer function with respect to the inner function, then multiply by the derivative of the inner function. Mathematically, this is written as:

• \( f'(x) = u'(v(x)) \times v'(x) \)

In our example problem, where \( f(x, y) = x^{2y} \), we apply the chain rule to differentiate with respect to \( x \). Here, \( f \) is composed of inner and outer functions involving \( x \) and \( y \). Understanding the chain rule is crucial for tackling such derivatives efficiently.

Exponential Functions

Exponential functions are functions of the form \( a^b \), where \( a \) is the base and \( b \) is the exponent. These functions frequently appear in various mathematical contexts due to their unique properties, such as constant proportional growth rates. In calculus, we often encounter the natural exponential function, \( e^x \), because of its simple derivative properties - its derivative is itself.
When dealing with functions like \( x^{2y} \), we express them in terms of \( e \) using the identity \( a^b = e^{b\ln a} \). This transformation enables us to apply the derivative rules of exponential functions, especially when partial derivatives are needed. Using exponential forms can simplify complex differentiation tasks, making it easier to find derivatives of expressions involving powers.

Partial Derivative with respect to x

Partial derivatives allow us to see how a multivariable function changes as one variable changes, while other variables are held constant. For the function \( f(x, y) = x^{2y} \), we start by finding the partial derivative with respect to \( x \). This means that we treat \( y \) as a constant.
Using the chain rule, the partial derivative \( \frac{\partial f}{\partial x} \) focuses on how the function changes with \( x \). In this case, it involves differentiating \( x^{2y} \) with respect to \( x \), considering \( 2y \) as a constant factor:

• \( \frac{\partial f}{\partial x} = (2y) x^{(2y-1)} \)

This expression shows how the function response to changes in \( x \) while \( y \) remains unchanged, effectively giving us the rate of change of \( f \) in the direction of the \( x \)-axis.

Partial Derivative with respect to y

Taking the partial derivative with respect to \( y \) tells us how the function changes as \( y \) varies, with \( x \) held constant. To differentiate \( f(x, y) = x^{2y} \) with respect to \( y \), rewrite it using our exponential identity: \( f(x, y) = e^{(\ln x)(2y)} \).
In this exponential form, we can apply the chain rule easily. Differentiate \((\ln x)(2y)\) with respect to \( y \):

• \( \frac{\partial((\ln x)(2y))}{\partial y} = 2\ln x \)

Then, apply this result into our differentiated exponential function:

• \( \frac{\partial f}{\partial y} = 2\ln x \times e^{(\ln x)(2y)} = 2\ln x \times x^{2y} \)

This partial derivative captures the influence of \( y \) on \( f(x, y) \), allowing us to observe how changes in \( y \) affect the function's behavior.