Problem 43 - Calculus Early Transcendentals

Question

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

Step-by-Step Solution

Answer

Question: Find the derivative of the function w(t) = x(t)y(t)z(t) with respect to t, where x(t) = 2t^4, y(t) = 3t^-1, and z(t) = 4t^-3. Answer: The derivative of w(t) with respect to t is -9t^-1.

Step 1: (Step 1: Find dx/dt, dy/dt, dz/dt)

(Differentiate the functions x, y, and z with respect to t. That is, find the derivatives dx/dt, dy/dt, and dz/dt. Using the power rule, which states that the derivative of t^n with respect to t is n*t^n-1, gives: dx/dt = 4*2*t^(4-1) = 8*t^3, dy/dt = -1*3*t^(-1-1) = -3*t^-2, and dz/dt = -3*4*t^(-3-1) = -12*t^-4, respectively. )

Step 2: (Step 2: Apply the chain rule to find dw/dt)

(Next, apply the chain rule to find the derivative dw/dt. The chain rule requires multiplying the derivative of the outer function by the derivative of the inner function. Given w = xyz, the derivative dw/dt can be found as follows: dw/dt = x (dy/dt z + y dz/dt) + dx/dt yz = 2t^4 (-3*t^-2 * 4t^-3 + 3t^-1 * -12t^-4) + 8t^3 * 3t^-1 * 4t^-3 = 2t^4 (-12t^-5 - 36t^-5) + 8t^3 * 12t^-4 = -t^-1 - 8t^-1 = -9t^-1 So, the derivative of w with respect to t is -9t^-1.)

Key Concepts

Chain Rule

The chain rule is a fundamental concept in calculus used to find the derivative of a composite function. A composite function is essentially a function within another function. For instance, in the exercise given, the function for derivative determination is composed of smaller functions: \( w = x y z \) where each involves \( t \), such as \( x = 2t^4 \), \( y = 3t^{-1} \), and \( z = 4t^{-3} \).

To apply the chain rule, you need to differentiate the outer function by multiplying it by the derivative of the inner functions. The key here is to deal with one function at a time while observing how these functions nest within each other. In practice, for a function expressed as \( w = f(g(t)) \), the derivative with respect to \( t \) is determined by \( \frac{dw}{dt} = \frac{dw}{dg} \times \frac{dg}{dt} \).

When working with multiple functions as in the problem, the chain rule helps in systematically breaking down each part to compute an overall derivative. It simplifies working with complex expressions, making the seemingly tough task manageable, one step at a time.

Power Rule

The power rule is one of the simplest yet most useful rules in calculus for finding derivatives. It's particularly handy when differentiating polynomial functions. The rule is straightforward: if \( u(t) = t^n \), then the derivative \( \frac{du}{dt} \) is \( n \cdot t^{n-1} \). This is applied directly to each term independently.

In the provided problem, the power rule is applied to each function \( x = 2t^4 \), \( y = 3t^{-1} \), and \( z = 4t^{-3} \):

• For \( x = 2t^4 \), \( \frac{dx}{dt} = 8t^3 \).
• For \( y = 3t^{-1} \), \( \frac{dy}{dt} = -3t^{-2} \).
• For \( z = 4t^{-3} \), \( \frac{dz}{dt} = -12t^{-4} \).

The utility of the power rule lies in its efficiency; it reduces the potential complexity of calculations by providing a fast way to determine derivatives without needing to revert to first principles every time. Understanding this rule well can speed up the process of solving many calculus problems.

Product Rule

The product rule comes into play when you have to differentiate a product of two functions. If you have two functions, say \( u(t) \) and \( v(t) \), then the derivative of their product \( u(t) \times v(t) \) with respect to \( t \) is: \( \frac{d}{dt}[u(t)v(t)] = u(t) \cdot \frac{d}{dt}[v(t)] + v(t) \cdot \frac{d}{dt}[u(t)] \).

In the given exercise, applying the product rule helps in deriving \( \frac{dw}{dt} \) where \( w = x y z \). Each of these components is a function of \( t \):

• Start by applying the rule to derivatives of each pair of products (like \( x \cdot dy/dt \cdot z \) etc.).

Ultimately, the rule assists in breaking down the derivative of a complex product into manageable parts, enabling you to handle derivatives of products of functions systematically and accurately. This technique, combined with previous rules, ensures a smooth path to comprehending the calculus derivative of products, especially in multi-variable contexts.