Problem 11 - Calculus for Biology and Medicine

Question

In Problems 1-16, find \(\partial f / \partial x\) and \(\partial f / \partial y\) for the given functions. $$ f(x, y)=e^{x} \sin (x y) $$

Step-by-Step Solution

Answer

\(\frac{\partial f}{\partial x} = e^x (\sin(xy) + y \cos(xy))\); \(\frac{\partial f}{\partial y} = xe^x \cos(xy)\).

Step 1: Differentiate with respect to x

To find the partial derivative \( \frac{\partial f}{\partial x} \), treat \( y \) as a constant. The function \( f(x, y) = e^x \sin(xy) \) can be differentiated using the product rule, which states if \( u(x) \) and \( v(x) \) are functions of \( x \), then \( \frac{d}{dx}(uv) = u'v + uv' \). Here, \( u(x) = e^x \) and \( v(x) = \sin(xy) \).

Step 2: Apply the product rule for differentiation

Differentiate each function: \( u'(x) = e^x \) (derivative of \( e^x \)) and \( v'(x) = y \cos(xy) \) (derivative of \( \sin(xy) \) with respect to \( x \)). Use the product rule: \[ \frac{\partial f}{\partial x} = e^x \sin(xy) + e^x(y \cos(xy)) \] Factoring out the common \( e^x \), we have: \[ \frac{\partial f}{\partial x} = e^x (\sin(xy) + y \cos(xy)) \]

Step 3: Differentiate with respect to y

Now, find \( \frac{\partial f}{\partial y} \) by treating \( x \) as a constant. Again apply the product rule: Differentiate \( u(y) = \sin(xy) \), and \( u'(y) = x \cos(xy) \). Thus, we have: \[ \frac{\partial f}{\partial y} = e^x \cdot x \cos(xy) \] \[ \frac{\partial f}{\partial y} = xe^x \cos(xy) \]

Key Concepts

Product Rule

The product rule is a fundamental rule in calculus used when differentiating products of two functions. In the context of partial derivatives, this rule is just as important. When you have a function posed as a product of two smaller functions, for example, like in our function \( f(x, y) = e^x \sin(xy) \), the product rule is your go-to tool for differentiation. The product rule formula is specified as \( \frac{d}{dx}(uv) = u'v + uv' \). This essentially means you take the derivative of the first function \( u \) while keeping the second function \( v \) unchanged, add it to the derivative of the second function \( v \) while keeping the first function \( u \) unchanged.

Here's a simplification:

• Differentiate \( u \) and multiply by \( v \)
• Add \( u \) multiplied by the derivative of \( v \)

This property is particularly useful in our function, where \( u(x) = e^x \) and \( v(x) = \sin(xy) \) are products, allowing us to simplify and organize our solution with ease.

Differentiation with respect to x

When differentiating a function with respect to a single variable, you hold the other variables constant. This process gives us the partial derivative with respect to that variable. In the given function \( f(x, y) = e^x \sin(xy) \), when differentiating with respect to \( x \), treat \( y \) as a constant.

During this differentiation, implement the product rule:

• Differentiate \( u(x) = e^x \) as \( u'(x) = e^x \)
• Differentiate \( v(x) = \sin(xy) \) with respect to \( x \). Here, since \( y \) is a constant, it becomes \( v'(x) = y \cos(xy) \)

Plug these results back into the product rule:
\( \frac{\partial f}{\partial x} = e^x \sin(xy) + e^x(y \cos(xy)) \).
Factor out the common term \( e^x \) to simplify the expression:
\( \frac{\partial f}{\partial x} = e^x (\sin(xy) + y \cos(xy)) \).
This simplification helps make it clear how each component contributes to the overall derivative.

Differentiation with respect to y

The process of differentiating \( f(x, y) \) with respect to \( y \) involves treating \( x \) as a constant. This technique is crucial for understanding how the function changes as \( y \) varies, while \( x \) stays fixed.

Here's how we handle \( f(x, y) = e^x \sin(xy) \):

• Differentiate the internal function \( \sin(xy) \) with respect to \( y \); here \( u(y) = \sin(xy) \) yields \( u'(y) = x \cos(xy) \), since \( x \) is constant
• The exponential portion \( e^x \) remains unchanged, as it does not contain \( y \)

Combine these to find the partial derivative:
\( \frac{\partial f}{\partial y} = e^x \cdot x \cos(xy) \).
Thus, the expression \( xe^x \cos(xy) \) shows how sensitive the function is to changes in \( y \), highlighting the interaction between \( x \), the exponential alert, and the trigonometric component. This provides a vivid illustration of the relationship between \( x \) and \( y \) in the function's behavior.