Problem 31 - Applied Mathematics: For the Managerial, Life, and Social Sciences

Question

In Exercises 31-34, suppose \(f\) and \(g\) are functions that are differentiable at \(x=1\) and that \(f(1)=2, f^{\prime}(1)=-1\), \(g(1)=-2\), and \(g^{\prime}(1)=3 .\) Find the value of \(h^{\prime}(1)\) \(h(x)=f(x) g(x)\)

Step-by-Step Solution

Answer

The derivative of the function \(h(x) = f(x)g(x)\) at \(x=1\) is \(h'(1) = 8\).

Step 1: Recall the Product Rule for Derivatives

The Product Rule states that if we have two functions \(u(x)\) and \(v(x)\), then the derivative of their product is given by: \[ (uv)' = u'v + uv' \] Now we have \(h(x) = f(x)g(x)\), so we will use the formula to find the derivative of \(h(x)\).

Step 2: Apply the Product Rule to h(x)

Using the Product Rule with \(u(x) = f(x)\) and \(v(x) = g(x)\), we obtain the derivative \(h'(x)\) as: \[ h'(x) = f'(x)g(x) + f(x)g'(x) \]

Step 3: Calculate h'(1)

Now that we have derived the expression for \(h'(x)\), we can plug in \(x=1\) and the given values for \(f(1), f'(1), g(1), g'(1)\) to find the value of \(h'(1)\): \[ h'(1) = f'(1)g(1) + f(1)g'(1) \] By plugging in the given values, we get: \[ h'(1) = (-1)(-2) + (2)(3) = 2+6 = 8 \] Thus, we have found the value of \(h'(1) = 8\).

Key Concepts

Differentiable Functions

The foundation of calculus is the concept of differentiability. A function is said to be differentiable at a point if it has a derivative at that point.

Think of the derivative as the rate at which the function's output is changing at any point along its curve. For a function to be differentiable, it must be continuous; there cannot be any sharp corners or discontinuities in its graph. When a function is differentiable, it's smooth and has a tangent line that gives the function's slope at any point.

A differentiable function can be visually imagined as a curve that you could smoothly draw without lifting your pencil off the paper. If you find a function difficult to draw because of jumps, breaks, or corners, these are indicators that the function might not be differentiable at those points.

Derivative of a Product

When dealing with the derivative of a product of two functions, simple rules of differentiation do not apply directly. That's where the product rule comes in as a critical strategy in calculus.

It's often assumed that the derivative of a product is the product of the derivatives, but this is not true. The correct approach is to apply the product rule, which provides a formula to find the derivative of the product of two differentiable functions. The essence of the product rule is to account for the way one function's rate of change affects the other and vice versa.

This rule is essential when solving problems that involve the multiplication of variable expressions, such as areas, volumes, or other quantities that depend on multiple variable factors.

Applying the Product Rule

The practical application of the product rule can be described through a methodical set of steps. To apply the product rule, follow these:

First, identify the two functions that are being multiplied. Let's call them function 'u' and function 'v'.
Then, differentiate both functions separately with respect to the variable of interest, usually 'x'. You get u' and v' as the derivatives.
Finally, apply the formula: the derivative of the product u×v is u'v + uv'.

This rule must be meticulously followed to correctly work through calculations involving the derivatives of products. As seen in the example, it was essential to correctly apply the product rule in order to find the value of the derivative of h at x=1. Without appropriately applying this rule, the resulting derivative would be incorrect, leading to a misunderstanding of how the function behaves at that point.