Problem 3 - Calculus with Concepts in Calculus

Question

Find the gradient of the function. $$ g(x, y)=e^{-2 x} \ln (y-4) $$

Step-by-Step Solution

Answer

The gradient is \( \nabla g(x, y) = \left( -2e^{-2x} \ln(y - 4), \frac{e^{-2x}}{y - 4} \right) \).

Step 1: Understand the Problem

The problem asks us to find the gradient of the function \( g(x, y) = e^{-2x} \ln(y - 4) \). The gradient of a function \( g(x, y) \) is a vector containing the partial derivatives of \( g \) with respect to each variable, \( x \) and \( y \).

Step 2: Find the Partial Derivative with respect to x

To find \( \frac{\partial g}{\partial x} \), treat \( y \) as a constant. The function \( g(x, y) = e^{-2x} \ln(y - 4) \) can be differentiated with respect to \( x \) using the product rule: if \( u = e^{-2x} \) and \( v = \ln(y - 4) \), then \( \frac{d}{dx}(uv) = u'v + uv' \). First, differentiate \( u = e^{-2x} \) to get \( u' = -2e^{-2x} \). Since \( v = \ln(y-4) \), its derivative \( v'\) with respect to \( x \) is zero because it doesn't contain \( x \). Thus, \( \frac{\partial g}{\partial x} = -2e^{-2x} \ln(y - 4) \).

Step 3: Find the Partial Derivative with respect to y

To find \( \frac{\partial g}{\partial y} \), treat \( x \) as a constant. The derivative of \( \ln(y - 4) \) with respect to \( y \) is \( \frac{1}{y - 4} \), while \( e^{-2x} \) is treated as a constant multiplier. Therefore, \( \frac{\partial g}{\partial y} = e^{-2x} \cdot \frac{1}{y - 4} \).

Step 4: Combine Partial Derivatives into the Gradient

Combine the partial derivatives into the gradient vector: the gradient of \( g(x, y) \) is \( abla g(x, y) = \left( \frac{\partial g}{\partial x}, \frac{\partial g}{\partial y} \right) = \left( -2e^{-2x} \ln(y - 4), \frac{e^{-2x}}{y - 4} \right) \).

Key Concepts

Partial derivatives

In multivariable calculus, a partial derivative of a function is the derivative with respect to one of its variables, holding the other variables constant. It is used to measure how the function changes as one specific variable changes, while others remain fixed. For the function given, \( g(x, y) = e^{-2x} \ln(y - 4) \), we find the partial derivatives with respect to \( x \) and \( y \).

• Partial Derivative with respect to \( x \): This is found by treating \( y \) as a constant. Using the product rule, the expression becomes \( \frac{\partial g}{\partial x} = -2e^{-2x} \ln(y - 4) \) as \( e^{-2x} \) is differentiable and \( \ln(y-4) \) acts as a constant.


• Partial Derivative with respect to \( y \): Here, \( x \) is constant. Differentiating \( \ln(y-4) \) gives \( \frac{1}{y - 4} \) and combining with the constant \( e^{-2x} \), we get \( \frac{\partial g}{\partial y} = e^{-2x} \cdot \frac{1}{y - 4} \).

Partial derivatives are foundational in understanding the behavior and changes in multivariable functions.

Multivariable calculus

Multivariable calculus extends the concepts of differentiation and integration to functions involving more than one variable. Instead of just one "slope", in multivariable calculus, we have gradients which are vectors showing the direction and rate of change of the function.

• Functions of Multiple Variables: Functions like \( g(x, y) = e^{-2x} \ln(y - 4) \) involve more than one variable, requiring a deeper understanding of how each variable affects the output independently and interactively.


• Gradient Vector: The gradient is a vector made up of all the partial derivatives of the function. For our function, the gradient is \( abla g(x, y) = \left( -2e^{-2x} \ln(y - 4), \frac{e^{-2x}}{y - 4} \right) \). This gradient vector tells us the rate and direction of the most significant change of the function.

Understanding multivariable calculus is crucial in fields like physics and engineering where systems vary with several parameters.

Product rule

The product rule is a fundamental tool in calculus used to find the derivative of the product of two functions. When dealing with functions that are products, like \( g(x, y) = e^{-2x} \ln(y - 4) \), the rule helps in differentiating effectively.

• Definition: If two functions are \( u(x) \) and \( v(x) \), their derivative using the product rule is \( (uv)' = u'v + uv' \). For our function, \( u = e^{-2x} \) and \( v = \ln(y - 4) \).


• Application: For the derivative with respect to \( x \), we need to differentiate \( e^{-2x} \) while treating \( \ln(y-4) \) as a constant. This gives us \( -2e^{-2x} \ln(y - 4) \) since the derivative of \( \ln(y-4) \) with respect to \( x \) is zero.

The product rule is especially useful when computing derivatives of functions that are not simply separable.