Problem 16 - Calculus Early Transcendentals

Question

Find the first partial derivatives of the following functions. $$f(s, t)=\frac{s-t}{s+t}$$

Step-by-Step Solution

Answer

The first partial derivatives are: $$\frac{\partial f}{\partial s}=\frac{2t}{(s+t)^2}$$ $$\frac{\partial f}{\partial t}=\frac{-2s}{(s+t)^2}$$

Step 1: Review function and differentiation rules

Our function is given as: $$f(s, t)=\frac{s-t}{s+t}$$ We will be using the Leibniz's Notation for finding partial derivatives of a function with respect to each independent variable (s and t). The rules of differentiation for addition, subtraction, and the division rule will be applied.

Step 2: Differentiate with respect to s

We will now find the partial derivative of the function with respect to s. Using the quotient rule of differentiation, we have: $$\frac{\partial f}{\partial s}=\frac{\frac{\partial}{\partial s} (s-t)\cdot(s+t)-\frac{\partial}{\partial s} (s+t)\cdot(s-t)}{(s+t)^2}$$ Now, let's differentiate each term inside the fractions. \(\frac{\partial}{\partial s}(s-t)=1\), \(\frac{\partial}{\partial s}(s+t)=1\): $$\frac{\partial f}{\partial s}=\frac{(1)\cdot(s+t)-(1)\cdot(s-t)}{(s+t)^2}$$ Now, we simplify the numerator: $$\frac{\partial f}{\partial s}=\frac{2t}{(s+t)^2}$$

Step 3: Differentiate with respect to t

We will now find the partial derivative of the function with respect to t. Using the quotient rule of differentiation, we have: $$\frac{\partial f}{\partial t}=\frac{\frac{\partial}{\partial t} (s-t)\cdot(s+t)-\frac{\partial}{\partial t} (s+t)\cdot(s-t)}{(s+t)^2}$$ Now, let's differentiate each term inside the fractions. \(\frac{\partial}{\partial t}(s-t)=-1\), \(\frac{\partial}{\partial t}(s+t)=1\): $$\frac{\partial f}{\partial t}=\frac{(-1)\cdot(s+t)-(1)\cdot(s-t)}{(s+t)^2}$$ Now, we simplify the numerator: $$\frac{\partial f}{\partial t}=\frac{-2s}{(s+t)^2}$$

Step 4: Write the results

Finally, we have both first-order partial derivatives: $$\frac{\partial f}{\partial s}=\frac{2t}{(s+t)^2}$$ $$\frac{\partial f}{\partial t}=\frac{-2s}{(s+t)^2}$$ These are the first partial derivatives of the given function \(f(s, t)=\frac{s-t}{s+t}\).