Problem 25 - Calculus Early Transcendentals

Question

Find the following derivatives. \(w_{s}\) and \(w_{t},\) where \(w=\frac{x-z}{y+z}, x=s+t, y=s t,\) and \(z=s-t\)

Step-by-Step Solution

Answer

Question: Compute the partial derivatives \(w_s\) and \(w_t\) of the function \(w = \frac{x-z}{y+z}\), given that x, y, and z are functions of s and t. Answer: The partial derivatives \(w_s\) and \(w_t\) of the function are: \(w_s = \frac{-t(x-z)+(x+z)}{(y+z)^2}\) \(w_t = \frac{s(x-z)+(x-z)}{(y+z)^2}\)

Step 1: Compute the partial derivatives of w with respect to x, y, and z.

Using the quotient rule, which states that \(\frac{d}{dt}\frac{u}{v}=\frac{vu'-u'v}{v^2}\), we can find the partial derivatives of w: \(\frac{\partial w}{\partial x} = \frac{-(y+z)}{(y+z)^2}\) \(\frac{\partial w}{\partial y} = \frac{x-z}{(y+z)^2}\) \(\frac{\partial w}{\partial z} = \frac{-(x+z)}{(y+z)^2}\)

Step 2: Compute the partial derivatives of x, y and z with respect to s and t.

Differentiate x, y, and z with respect to s and t: \(\frac{\partial x}{\partial s} = 1\) \(\frac{\partial x}{\partial t} = 1\) \(\frac{\partial y}{\partial s} = t\) \(\frac{\partial y}{\partial t} = s\) \(\frac{\partial z}{\partial s} = 1\) \(\frac{\partial z}{\partial t} = -1\)

Step 3: Use the chain rule to find the derivatives \(w_s\) and \(w_t\)

To find \(w_s\) and \(w_t\), we use the chain rule: \(w_s = \frac{\partial w}{\partial x}\frac{\partial x}{\partial s} + \frac{\partial w}{\partial y}\frac{\partial y}{\partial s} + \frac{\partial w}{\partial z}\frac{\partial z}{\partial s}\) \(w_t = \frac{\partial w}{\partial x}\frac{\partial x}{\partial t} + \frac{\partial w}{\partial y}\frac{\partial y}{\partial t} + \frac{\partial w}{\partial z}\frac{\partial z}{\partial t}\) Compute \(w_s\) and \(w_t\): \(w_s = \left(-\frac{y+z}{(y+z)^2}\right)(1) + \left(\frac{x-z}{(y+z)^2}\right)(t) + \left(-\frac{x+z}{(y+z)^2}\right)(1)\) \(w_s = \frac{-t(x-z)+(x+z)}{(y+z)^2}\) \(w_t = \left(-\frac{y+z}{(y+z)^2}\right)(1) + \left(\frac{x-z}{(y+z)^2}\right)(s) + \left(-\frac{x+z}{(y+z)^2}\right)(-1)\) \(w_t = \frac{s(x-z)+(x-z)}{(y+z)^2}\)