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Chapter 3: Problem 17

Find the derivative of the given function. $$ f(x)=\sqrt{9+\sqrt{9-x}} $$

Short Answer

Expert verified
The derivative of the function is \( f'(x) = -\frac{(9-x)^{-1/2}}{2(9 + \sqrt{9-x})^{1/2}} \).

Step by step solution

01

Rewrite the Function

Express the function in a more convenient form for differentiation. Start by writing the given function as \( f(x) = (9 + \sqrt{9 - x})^{1/2} \).
02

Apply the Chain Rule

The outer function is \( u^{1/2} \) where \( u = 9 + \sqrt{9 - x} \). Using the chain rule, \( f'(x) = \frac{1}{2} u^{-1/2} \cdot u' \).
03

Differentiate Inner Function

Differentiate \( u = 9 + \sqrt{9 - x} \). The derivative of 9 is 0, so we only need to differentiate \( \sqrt{9 - x} \). This gives \( u' = (9 - x)^{-1/2} \cdot (-1) \).
04

Simplify the Derivative

Combine the results from the chain rule application. Substitute \( u \) and \( u' \) into the chain rule formula: \( f'(x) = \frac{1}{2}(9 + \sqrt{9 - x})^{-1/2} \cdot (9 - x)^{-1/2}(-1) \).
05

Final Expression

Simplify the expression for the final derivative: \( f'(x) = -\frac{(9-x)^{-1/2}}{2(9 + \sqrt{9 - x})^{1/2}} \).

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Key Concepts

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

Chain Rule
When we talk about the chain rule in calculus, we refer to a method used to differentiate composite functions. A composite function is a function within another function.

Imagine you have a function that looks like this: f(g(x)). The chain rule tells us how to take the derivative of such a function. Specifically, the derivative of f(g(x)) is the derivative of the outer function evaluated at the inner function times the derivative of the inner function.

In general terms, if u = g(x) and y = f(u), then using the chain rule:

  • The derivative of y with respect to x is (dy/du) * (du/dx)
For example, in our exercise, we simplified the function and recognized that within (9 + sqrt(9 - x))^(1/2), the outside function is (... )^(1/2) and the inside function is 9 + sqrt(9 - x). This is the core idea behind the chain rule.
Differentiation
Differentiation is another fundamental concept in calculus. It refers to the process of finding the derivative of a function. The derivative measures how a function's value changes as its input changes.

To differentiate a function, we apply specific rules that depend on the form of the function. Common rules include:

  • The power rule: For any function of the form x^n, the derivative is n*x^(n-1).
  • The sum rule: The derivative of a sum of functions is the sum of their derivatives.
  • The chain rule: As mentioned before, it's used for composite functions.

For the function in our exercise, we first needed to rewrite it in a suitable form. We then applied the chain rule to differentiate it step-by-step. Understanding differentiation and its rules allows you to tackle a wide range of problems.
Calculus
Calculus is the branch of mathematics that deals with change. It includes two main parts: differential calculus and integral calculus.

  • Differential calculus deals with the concept of a derivative, representing the rate of change of a quantity.
  • Integral calculus, on the other hand, deals with the concept of an integral, representing the accumulation of quantities.

The problem we solved here falls under differential calculus because we were finding the derivative of the given function. The ability to find derivatives is crucial in physics, engineering, economics, and many other fields. It helps describe how quantities change over time, optimize functions, and model real-world phenomena.

By mastering the techniques of differentiation, such as applying the chain rule, you are well on your way to solving complex problems in calculus.

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Most popular questions from this chapter

In Exercises 9 through 12 , the motion of a particle is along a horizontal line according to the given equation of motion, where \(s \mathrm{ft}\) is the directed distance of the particle from a point \(O\) at \(t \mathrm{sec}\). The positive direction is to the right. Determine the intervals of time when the particle is moving to the right and when it is moving to the left. Also determine when the particle reverses its direction. Show the behavior of the motion by a figure similar to Fig. 3.2.2, choosing values of \(t\) at random but including the values of \(t\) when the particle reverses its direction. $$ s=t^{3}+3 t^{2}-9 t+4 $$

Find the slope of the tangent line to the graph at the point \(\left(x_{1}, y_{1}\right) .\) Make a table of values of \(x, y\), and \(m\) at various points on the graph, and include in the table all points where the graph has a horizontal tangent. Draw a sketch of the graph. $$ y=x^{2}-6 x+9 $$

Find the derivative of the given function. $$ H(z)=\left(z^{3}-3 z^{2}+1\right)^{-3} $$

Do each of the following: (a) Draw a sketch of the graph of the function; (b) determine if \(f\) is continuous at \(x_{1} ;\) (c) find \(f^{\prime}-\left(x_{1}\right)\) and \(f_{+}^{\prime}\left(x_{1}\right)\) if they exist; (d) determine if \(f\) is differentiable at \(x_{1}\). $$ f(x)=\sqrt[3]{x+1} ; x_{1}=-1 $$

An object is moving along a straight line according to the equation of motion \(s=3 t /\left(t^{2}+9\right)\), with \(t \geq 0\), where \(s\) ft is the directed distance of the object from the starting point at \(t \mathrm{sec}\). (a) What is the instantaneous velocity of the object at \(t_{1}\) sec? (b) What is the instantaneous velocity at \(1 \mathrm{sec}\) ? (c) At what time is the instantaneous velocity zero?
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