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Chapter 12: Problem 43

Find the indicated derivative for the following functions. $$d w / d t, \text { where } w=x y z, x=2 t^{4}, y=3 t^{-1}, \text {and } z=4 t^{-3}$$

Short Answer

Expert verified
Question: Find the derivative of w = xyz with respect to t, given x = 2t^4, y = 3t^(-1), and z = 4t^(-3). Answer: The derivative of w with respect to t is -72t^(-1).

Step by step solution

01

Find partial derivatives of w with respect to x, y, and z

Begin by finding the partial derivatives of the function \(w=xyz\) w.r.t. \(x\), \(y\), and \(z\). For each one, treat the other two variables as constants. $$\frac{\partial w}{\partial x} = yz$$ $$\frac{\partial w}{\partial y} = xz$$ $$\frac{\partial w}{\partial z} = xy$$
02

Determine the derivatives of x, y, and z w.r.t. t

In order to apply the chain rule, we need to find the derivatives of \(x\), \(y\), and \(z\) w.r.t. \(t\). Given, \(x = 2t^4\), \(y = 3t^{-1}\), and \(z = 4t^{-3}\). $$\frac{dx}{dt} = 8t^3$$ $$\frac{dy}{dt} = -3t^{-2}$$ $$\frac{dz}{dt} = -12t^{-4}$$
03

Apply the chain rule to find dw/dt

Now, use the chain rule to find \(\frac{dw}{dt}\). This is accomplished by multiplying each of the partial derivatives by their respective total derivatives w.r.t. \(t\) and summing them. $$\frac{dw}{dt} = \frac{\partial w}{\partial x} \frac{dx}{dt} + \frac{\partial w}{\partial y} \frac{dy}{dt} + \frac{\partial w}{\partial z} \frac{dz}{dt}$$ Substitute the values we found in steps 1 and 2: $$\frac{dw}{dt} = yz(8t^3) + xz(-3t^{-2}) + xy(-12t^{-4})$$ Replace \(x\), \(y\), and \(z\) with their expressions in terms of \(t\): $$\frac{dw}{dt} = (3t^{-1})(4t^{-3})(8t^3) + (2t^4)(4t^{-3})(-3t^{-2}) + (2t^4)(3t^{-1})(-12t^{-4})$$
04

Simplify the expression

Simplify the expression for \(\frac{dw}{dt}\): $$\frac{dw}{dt} = 24t^{-1} - 24t^{-1} - 72t^{-1}$$ Combine like terms: $$\frac{dw}{dt} = -72t^{-1}$$ Thus, the indicated derivative is: $$\frac{dw}{dt} = -72t^{-1}$$

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