Problem 10 - Calculus Early Transcendentals: Pearson New International Edition

Question

Find all first partial derivatives of each function. \(f(s, t)=\ln \left(s^{2}-t^{2}\right)\)

Step-by-Step Solution

Answer

The partial derivatives are \( \frac{2s}{s^2 - t^2} \) with respect to \(s\) and \( \frac{-2t}{s^2 - t^2} \) with respect to \(t\).

Step 1: Identify the Need for Partial Derivatives

To find partial derivatives, we differentiate the function with respect to each of its variables separately (keeping the other variable constant). This way, we derive the expressions for the slope of the function in regard to each variable.

Step 2: Differentiate with Respect to s

To find the partial derivative of the function with respect to \(s\), treat \(t\) as a constant. Apply the chain rule where the outer function is the natural logarithm and the inner function is \(s^2 - t^2\).The derivative of \( \ln(u) \) is \( \frac{1}{u} \) and the derivative of \( s^2 - t^2 \) with respect to \( s \) is \( 2s \).So, the partial derivative \(\frac{\partial f}{\partial s}\) is:\[ \frac{\partial f}{\partial s} = \frac{1}{s^2 - t^2} \cdot 2s = \frac{2s}{s^2 - t^2} \]

Step 3: Differentiate with Respect to t

To find the partial derivative of the function with respect to \(t\), treat \(s\) as a constant. Again, use the chain rule where the outer function is the natural logarithm and the inner function remains \(s^2 - t^2\).The derivative of \( \ln(u) \) is \( \frac{1}{u} \), and the derivative of \( s^2 - t^2 \) with respect to \( t \) is \(-2t\).Thus, the partial derivative \(\frac{\partial f}{\partial t}\) is:\[ \frac{\partial f}{\partial t} = \frac{1}{s^2 - t^2} \cdot (-2t) = \frac{-2t}{s^2 - t^2} \]

Key Concepts

Chain Rule in Differentiation

The chain rule is a crucial technique in calculus used to differentiate composite functions, that is, when you have a function inside another function. When applying the chain rule, you essentially take the derivative of the outer function and multiply it by the derivative of the inner function. This is especially important when dealing with functions like our exercise's example, where we have the natural logarithm of an expression involving both variables.
To visualize:

• Outer function: natural logarithm, denoted as \(\ln(u)\)
• Inner function: the expression \(s^2 - t^2\)

When differentiating such a function, you find the derivative by:

• First, differentiating the outer function based on the inner: \( \frac{d}{du} [\ln(u)] = \frac{1}{u} \)
• Next, multiplying by the derivative of the inner function with respect to the variable of interest.

Thus, for partial derivatives in multivariable functions, the chain rule helps us find how one variable affects the function independently of others.

Natural Logarithm Differentiation

Differentiating a natural logarithm is distinct because of its base, \(e\), the natural exponential constant. The derivative of \(\ln(x)\) with respect to \(x\) is \(\frac{1}{x}\). This simple rule becomes slightly more complex when the argument of the logarithm is itself a function, as seen in our example with \(\ln(s^2 - t^2)\).
During differentiation:

• You first address the logarithm's output by applying \(\frac{1}{u}\), where \(u\) is the argument \(s^2 - t^2\).
• The effectiveness of chain rule coupling comes back into play as you then differentiate \(u\) with respect to each variable individually: \(2s\) for \(s\) and \(-2t\) for \(t\).

By combining these efforts, the differentiation caters to the composite nature of the function, ensuring each variable's influence is properly accounted for.

Partial Derivatives and Multivariable Calculus

Partial derivatives are foundational in multivariable calculus. They represent the rate of change of a function with respect to one variable while keeping others constant. For multivariable functions like \(f(s, t) = \ln(s^2 - t^2)\), each variable contributes uniquely to the function's behavior. Understanding each partial derivative tells us precisely how changes in each independent variable affect the function.
The practical steps include identifying each variable in the context of other constants:

• Keeping \(t\) constant, find \(\frac{\partial f}{\partial s}\) to express how changes in \(s\) alone affect \(f\).
• Conversely, by keeping \(s\) constant, determine \(\frac{\partial f}{\partial t}\) to reflect \(f\)'s responsiveness to \(t\).

In multivariable calculus, these computations are integral to more advanced topics such as gradient determination, optimization problems, and understanding surface behaviors based on variable shifts.