Problem 33 - Applied Mathematics: For the Managerial, Life, and Social Sciences

Question

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 \(h^{\prime}(1)\) \(h(x)=\frac{x f(x)}{x+g(x)}\)

Step-by-Step Solution

Answer

The derivative of the function \(h(x) = \frac{x f(x)}{x+g(x)}\) with respect to \(x\) at \(x=1\) is \(h'(1) = -1\).

Step 1: Find \(u'(x)\) and \(v'(x)\)

Using the product rule, calculate the derivative of \(u(x) = xf(x)\): \(u'(x) = f(x) + xf'(x)\) Now find the derivative of \(v(x) = x + g(x)\): \(v'(x) = 1 + g'(x)\)

Step 2: Compute \(h'(x)\) using the quotient rule

The quotient rule tells us that: \(h'(x) = \frac{u'(x)v(x) - u(x)v'(x)}{v(x)^2}\) Substitute the expressions for \(u(x)\), \(v(x)\), \(u'(x)\), and \(v'(x)\): \(h'(x) = \frac{(f(x) + xf'(x))(x + g(x)) - (xf(x))(1 + g'(x))}{(x+g(x))^2}\)

Step 3: Substitute the given values and find \(h'(1)\)

Plug in the values of \(f(1)\), \(f'(1)\), \(g(1)\), and \(g'(1)\) into the expression for \(h'(x)\): \(h'(1) = \frac{(2 - 1)(1 - 2) - (1)(-1 + 3)}{(1 - 2)^2}\) Further simplify the expression: \(h'(1) = \frac{1(1) - (1)(2)}{(-1)^2}\) Calculate the final answer: \(h'(1) = \frac{-1}{1}\) Therefore, \(h'(1) = -1\).

Key Concepts

Derivatives

The foundation of calculus lies in the concept of derivatives. A derivative is a measure of how a function's output value changes as its input value changes. In simpler terms, it tells us the rate of change or the slope of the function at any point. The basic notation for the derivative of a function f(x) with respect to x is f'(x) or \(\frac{df}{dx}\).

• For a constant function f(x) = c, where c is a constant, the derivative is zero because a constant function doesn’t change as x changes.
• For a linear function f(x) = mx + b, the derivative is the constant m because the rate of change is constant.

Understanding derivatives is crucial because they form the basis for more complex rules in calculus, including the product rule, quotient rule, and chain rule, which are methods for finding derivatives of more complicated functions.

Product Rule

The product rule is a formula used to find the derivative of a product of two functions. The rule states that if we have two differentiable functions, f(x) and g(x), then the derivative of their product h(x) = f(x)\cdot g(x) is given by:

\[h'(x) = f(x)g'(x) + f'(x)g(x)\]

Essentially, you take the derivative of the first function and multiply it by the second function as it is, and then add the product of the first function as it is with the derivative of the second function. This rule allows us to easily differentiate functions that are multiplied together without having to resort to more complex algebraic manipulation.

Quotient Rule

When dealing with the division of two functions, the quotient rule comes into play. It is a method for finding the derivative of a quotient of two functions. For functions f(x) and g(x), where both are differentiable and g(x) is not zero, the derivative of the quotient h(x) = \(\frac{f(x)}{g(x)}\) is:

\[h'(x) = \frac{f'(x)g(x) - f(x)g'(x)}{g(x)^2}\]

This rule can be understood as taking the denominator times the derivative of the numerator, minus the numerator times the derivative of the denominator, all over the square of the denominator. It's important to note the switch of the subtraction that takes place - a common point where students may make mistakes.

Chain Rule

Complex functions often comprise of compositions of multiple functions. To differentiate these composite functions, we use the chain rule. The chain rule states that if you have two functions f(g(x)), where f(x) and g(x) are differentiable, then the derivative of the composite function is:

\[\frac{d}{dx}[f(g(x))] = f'(g(x))\cdot g'(x)\]

This means you first take the derivative of the outer function with respect to the inner function (f'(g(x))) and then multiply it by the derivative of the inner function (g'(x)). The chain rule is especially powerful as it allows us to tackle a wide range of functions by breaking them down into their component parts.

For students, remembering to multiply by the derivative of the inner function is critical, as it’s easy to overlook this step in the process.