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Chapter 9: Problem 33

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

Short Answer

Expert verified
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 by step solution

01

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 \(v(x) = x + g(x)\): \(v'(x) = 1 + g'(x)\)
02

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 \(v'(x)\): \(h'(x) = \frac{(f(x) + xf'(x))(x + g(x)) - (xf(x))(1 + g'(x))}{(x+g(x))^2}\)
03

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, \(h'(1) = -1\).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

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鈥檛 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鈥檚 easy to overlook this step in the process.

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Most popular questions from this chapter

Find the derivative of each function. \(f(x)=\sqrt{3 x-2}\)

The body mass index (BMI) measures body weight in relation to height. A BMI of 25 to \(29.9\) is considered overweight, a BMI of 30 or more is considered obese, and a BMI of 40 or more is morbidly obese. The percent of the U.S. population that is obese is approximated by the function \(P(t)=0.0004 t^{3}+0.0036 t^{2}+0.8 t+12 \quad(0 \leq t \leq 13)\) where \(t\) is measured in years, with \(t=0\) corresponding to the beginning of 1991 . Show that the rate of the rate of change of the percent of the U.S. population that is deemed obese was positive from 1991 to 2004 . What does this mean?

A study on formaldehyde levels in 900 homes indicates that emissions of various chemicals can decrease over time. The formaldehyde level (parts per million) in an average home in the study is given by $$ f(t)=\frac{0.055 t+0.26}{t+2} \quad(0 \leq t \leq 12) $$ where \(t\) is the age of the house in years. How fast is the formaldehyde level of the average house dropping when it is new? At the beginning of its fourth year?

In Section \(9.4\), we proved that $$ \frac{d}{d x}\left(x^{n}\right)=n x^{n-1} $$ for the special case when \(n=2\). Use the chain rule to show that $$ \frac{d}{d x}\left(x^{1 / n}\right)=\frac{1}{n} x^{1 / n-1} $$ for any nonzero integer \(n\), assuming that \(f(x)=x^{1 / n}\) is differentiable. Hint: Let \(f(x)=x^{1 / n}\) so that \([f(x)]^{n}=x\). Differentiate both sides with respect to \(x\).

Find the derivative of each function. \(f(t)=\frac{1}{\sqrt{2 t-3}}\)
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