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Find \(\frac{d y}{d x}\) for each pair of functions. $$ y=\sqrt{7-3 u}, \text { where } u=x^{2}-9 $$

Short Answer

Expert verified
\(\frac{dy}{dx} = \frac{-3x}{\sqrt{34-3x^2}}\)

Step by step solution

01

Identify the Inner Function

Given the function \(u = x^2 - 9\), identify \(u\) as the inner function.
02

Differentiate the Inner Function

Differentiate \(u = x^2 - 9\) with respect to \(x\). This gives \(\frac{du}{dx} = 2x\).
03

Identify the Outer Function

Given the function \(y = \sqrt{7 - 3u}\), identify \(y\) as the outer function, where the argument is \(7 - 3u\).
04

Differentiate the Outer Function

Differentiate \(y = \sqrt{7 - 3u}\) with respect to \(u\). This gives \(\frac{dy}{du} = \frac{1}{2}\cdot(7-3u)^{-1/2} \cdot (-3)\).
05

Introduce the Chain Rule

Apply the Chain Rule for differentiation: \(\frac{dy}{dx} = \frac{dy}{du} \cdot \frac{du}{dx}\).
06

Calculate \(\frac{dy}{dx}\)

Substitute the derivatives: \(\frac{dy}{dx} = \left(\frac{-3}{2\sqrt{7-3u}}\right) \cdot (2x)\). Simplify to get \(\frac{dy}{dx} = \frac{-3x}{\sqrt{7-3u}}\).
07

Substitute Back \(u\)

Substitute \(u = x^2 - 9\) back into the expression: \(\frac{dy}{dx} = \frac{-3x}{\sqrt{7-3(x^2 - 9)}}\), which simplifies to \(\frac{-3x}{\sqrt{34-3x^2}}\).

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Key Concepts

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

Chain Rule
When dealing with functions known as composite functions, the chain rule becomes incredibly useful. The chain rule is a fundamental theorem in calculus that allows us to differentiate a composite function. Simply put, if you have two functions, say, an outer function and an inner function, the chain rule helps in finding the derivative of their composite.
Here’s the essential bit:
  • Identify the inner function, which is the function inside another function.
  • Identify the outer function, the surrounding layer.
Once you have these, the chain rule states: \[\frac{dy}{dx} = \frac{dy}{du} \cdot \frac{du}{dx}\]
What this means is to multiply the derivative of the outer function relative to the inner with that of the inner function.
This allows breaking down the complexity of differentiating layered functions piece by piece and then reconstruct them with their derivatives neatly.
Derivatives
The derivative of a function measures how the function's value changes as its input changes. It's the cornerstone of calculus, expressing the rate of change. Think of it like speed, which is a measure of how distance changes over time.
Here's a simple breakdown on differentiating:
  • Direct derivatives: These are straightforward and involve applying standard differentiation rules to basic functions like polynomial, exponential, etc.
  • Indirect or composite derivatives: Here, rules like the chain rule come into play, as seen in our exercise.
In the example, \[ \frac{du}{dx} = 2x \] was found by differentiating the simple polynomial, \( x^2 - 9 \). This simple derivative becomes a component of our chain rule application. Derivatives not only tell us about the slope of a line at any point on a curve but also about maximum and minimum values of functions. Understanding derivatives means understanding how a function behaves, curves, and moves.
Composite Functions
Composite functions can be tricky to untangle at first. They are functions formed by combining two or more functions, like nested layers of an onion. Here’s what to keep in mind:
  • Inner and outer function: Recognizing which is nested in which is crucial.
  • Application of rules: The chain rule is a primary tool for differentiation of composites.
In our exercise, \( y = \sqrt{7-3u} \) is made up of two functions: the square root (outer) and \( 7 - 3u \) (inner).
A composite occurs when you replace the variable in one function with another function. Thus, working with composite functions often requires dealing with layers of differentiation.
Start by tackling each layer separately using their respective rules, then use the chain rule to bring it all together. This kind of step-by-step dissection helps keep complex problems manageable and less daunting.

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Most popular questions from this chapter

Below are the steps in the simplification of the difference quotient for \(f(x)=\sqrt{x}\) (see Example 8). Provide a brief justification for each step. \(\frac{f(x+h)-f(x)}{h}=\frac{\sqrt{x+h}-\sqrt{x}}{h}\) a) \(=\frac{\sqrt{x+h}-\sqrt{x}}{h} \cdot\left(\frac{\sqrt{x+h}+\sqrt{x}}{\sqrt{x+h}+\sqrt{x}}\right)\) b) \(=\frac{x+h+\sqrt{x} \sqrt{x+h}-\sqrt{x} \sqrt{x+h}-x}{h(\sqrt{x+h}+\sqrt{x})}\) c) \(=\frac{x+h-x}{h(\sqrt{x+h}+\sqrt{x})}\) d) \(=\frac{h}{h(\sqrt{x+h}+\sqrt{x})}\) e) \(=\frac{1}{\sqrt{x+h}+\sqrt{x}}\)

Population growth rate. In \(t\) years, the population of Kingsville grows from 100,000 to a size \(P\) given by \(P(t)=100,000+2000 t^{2}\) a) Find the growth rate, \(d P / d t\). b) Find the population after 10 yr. c) Find the growth rate at \(t=10\). d) Explain the meaning of your answer to part (c).

Let \(f\) and \(g\) be differentiable over an open interval containing \(x=a\). If $$ \lim _{x \rightarrow a}\left(\frac{f(x)}{g(x)}\right)=\frac{0}{0} \quad \text { or } \quad \lim _{x \rightarrow a}\left(\frac{f(x)}{g(x)}\right)=\frac{\pm \infty}{\pm \infty} $$ and if \(\lim _{x \rightarrow a}\left(\frac{f^{\prime}(x)}{g^{\prime}(x)}\right)\) exists, then $$ \lim _{x \rightarrow a}\left(\frac{f(x)}{g(x)}\right)=\lim _{x \rightarrow a}\left(\frac{f^{\prime}(x)}{g^{\prime}(x)}\right) $$ The forms \(0 / 0\) and \(\pm \infty / \pm \infty\) are said to be indeterminate. In such cases, the limit may exist, and l'Hôpital's Rule offers a way to find the limit using differentiation. For example, in Example 1 of Section \(1.1,\) we showed that $$ \lim _{x \rightarrow 1}\left(\frac{x^{2}-1}{x-1}\right)=2 $$ Since, for \(x=1,\) we have \(\left(x^{2}-1\right)(x-1)=0 / 0,\) we differentiate the numerator and denominator separately, and reevaluate the limit: $$ \lim _{x \rightarrow 1}\left(\frac{x^{2}-1}{x-1}\right)=\lim _{x \rightarrow 1}\left(\frac{2 x}{1}\right)=2 $$ Use this method to find the following limits. Be sure to check that the initial substitution results in an indeterminate form. $$ \lim _{x \rightarrow-4}\left(\frac{x^{2}+x-12}{x+4}\right) $$

In New York City, taxicabs charge passengers \(\$ 2.50\) for entering a cab and then \(\$ 0.50\) for each one-fifth of a mile (or fraction thereof) traveled. (There are additional charges for slow traffic and idle times, but these are not considered in this problem.) If \(x\) represents the distance traveled in miles, then \(C(x)\) is the cost of the taxi fare, where $$ \begin{array}{ll} C(x)=\$ 2.50, & \text { if } x=0 \\ C(x)=\$ 3.00, & \text { if } 0 < x \leq 0.2 \\ C(x)=\$ 3.50, & \text { if } 0.2 < x \leq 0.4, \\ C(x)=\$ 4.00, & \text { if } 0.4 < x \leq 0.6, \end{array} $$ and so on. The graph of C is shown below. Using the graph of the taxicab fare function, find each of the following limits, if it exists. $$ \lim _{x \rightarrow 0.6^{-}} C(x), \lim _{x \rightarrow 0.6^{+}} C(x), \lim _{x \rightarrow 0.6} C(x) $$

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