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Differentiate implicily to find \(d y / d x\). $$ y^{3}=x^{5} $$

Short Answer

Expert verified
\( \frac{dy}{dx} = \frac{5x^4}{3y^2} \)

Step by step solution

01

Differentiate Both Sides with Respect to x

Apply the differentiation operator to both sides of the equation. Remember that you need to use the chain rule when differentiating the left-hand side because it involves the variable y. \[ \frac{d}{dx} (y^3) = \frac{d}{dx} (x^5) \]
02

Apply the Chain Rule on the Left-Hand Side

For the left-hand side, apply the chain rule. The outside function is y cubed, and the inside function is y, so the derivative of y cubed is 3y虏 times the derivative of y (dy/dx). \[ \frac{d}{dx} (y^3) = 3y^2 \frac{dy}{dx} \]
03

Differentiate the Right-Hand Side

On the right-hand side, differentiate x^5 with respect to x. \[ \frac{d}{dx} (x^5) = 5x^4 \]
04

Combine Results

Now combine the differentiated results from both sides into a single equation:\[ 3y^2 \frac{dy}{dx} = 5x^4 \]
05

Solve for dy/dx

To solve for \( \frac{dy}{dx} \), rearrange the equation by dividing both sides by 3y虏:\[ \frac{dy}{dx} = \frac{5x^4}{3y^2} \]

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Key Concepts

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

Implicit Differentiation
Implicit differentiation is a technique used when we have an equation involving both x and y, and we cannot easily solve for y in terms of x. Instead of explicitly finding y, we differentiate both sides of the equation with respect to x. This method often involves using the chain rule, especially when y is raised to a power or appears in a complex function. Understanding implicit differentiation is crucial for handling equations where y isn't isolated on one side. Additionally, make sure to always apply the chain rule correctly to differentiate terms involving y.
Chain Rule
The chain rule is a fundamental concept in calculus used to differentiate composite functions. It states that if a variable y is a function of u, which is itself a function of x, then the derivative of y with respect to x can be found by multiplying the derivative of y with respect to u by the derivative of u with respect to x.
In our exercise, where we have an equation involving y鲁, we use the chain rule to handle the differentiation. The outside function is the power function y鲁, and the inside function is y. So we first differentiate y鲁 with respect to y, getting 3y虏, then multiply by the derivative of y with respect to x (dy/dx). This gives us:
\( \frac{d}{dx} (y^3) = 3y^2 \frac{dy}{dx} \).
Differentiation
Differentiation is the process of finding the derivative of a function. The derivative represents the rate at which a function is changing at any given point. It's a crucial tool in calculus for understanding behaviors of functions, like slopes and rates of change.
For the given exercise, we needed to differentiate both sides of the equation y鲁 = x鈦 with respect to x. First, we applied differentiation to each side:
\( \frac{d}{dx} (y^3) = \frac{d}{dx} (x^5) \)
Then, using the power rule for x鈦, we obtained 5x鈦 for the right-hand side. For the left-hand side, we applied the chain rule as discussed earlier.
dy/dx
dy/dx represents the derivative of y with respect to x. In other words, it tells us how y changes as x changes. When dealing with implicit differentiation, dy/dx often shows up as a crucial part of the process.
In our example, after differentiating both sides of the equation using the chain rule and power rule, we obtained the expression involving dy/dx:
\( 3y^2 \frac{dy}{dx} = 5x^4 \)
To solve for dy/dx, we rearranged the equation to isolate dy/dx on one side:
\( \frac{dy}{dx} = \frac{5x^4}{3y^2} \)
This final step gives us the rate at which y changes with respect to x, now expressed explicitly.
Implicit Function
An implicit function is a function where y is defined implicitly in terms of x rather than explicitly. Instead of having y = f(x), we have a relationship involving both x and y, like in the equation y鲁 = x鈦. In cases like this, finding the derivative directly is not feasible, so we use implicit differentiation.
By differentiating implicitly, we treat y as an implicit function of x. This approach allows us to handle more complex relationships between x and y without needing to solve for y first. The process we followed in the exercise demonstrates how implicit functions and implicit differentiation work together.

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Most popular questions from this chapter

Differentiate implicily to find \(d y / d x\). $$ y^{2}=\frac{x^{2}-1}{x^{2}+1} $$

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When we speak, the position of the tongue changes the shapes of the mouth and throat cavity, thereby changing the sound that is generated. For example, to pronounce the first vowel of the word father, the mouth is much wider than the throat cavity. For many vowel sounds, the vocal tract may be modeled by two adjoining tubes, with one end closed (the glottis) and the other end open (the lips). We denote by \(A_{m}\) and \(L_{m}\) the cross-sectional area and length of the mouth, and we denote by \(A_{t}\) and \(L_{t}\) the cross-sectional area and length of the throat cavity. The shape of the vocal tract tends to promote certain sound frequencies. \({ }^{15}\) Let \(c\) be the speed of sound. If \(\int\) is a frequency promoted by the vocal tract (in Hertz) and we let \(x=2 \pi \int / c\), then \(x\) is a solution of $$ \frac{1}{A_{m}} \tan \left(L_{m} x\right)-\frac{1}{A_{t}} \cot \left(L_{t} x\right)=0 $$ a) For a certain speaker of the first vowel of the word father; \(A_{m}=10 A_{t}\) (that is, the mouth opening is 10 times larger than the throat opening),\(L_{m}=8 \mathrm{~cm}\), and \(L_{t}=9.7 \mathrm{~cm} .^{16}\) Use this information to show that $$ \tan (8 x)-10 \cot (9.7 x)=0 $$ a) For a certain speaker of the first vowel of the word father; \(A_{m}=10 A_{t}\) (that is, the mouth opening is 10 times larger than the throat opening),\(L_{m}=8 \mathrm{~cm}\), and \(L_{t}=9.7 \mathrm{~cm} .^{16}\) Use this information to show that $$ \tan (8 x)-10 \cot (9.7 x)=0 $$ b) Use a grapher to sketch the graph of \(y=\tan (8 x)-10 \cot (9.7 x)\) using the graphing window \([0,0.5,-2,2] .\) From the graph, estimate the first three \(x\) -intercepts. c) Use Newton's method to find the first three solutions. Use \(x_{1}=0.1, x_{1}=0.2\), and \(x_{1}=0.45\) as your three starting points. d) The speed of sound is approximately \(35,400 \mathrm{~cm} / \mathrm{s}\). Use this value for \(c\) and the relationship \(x=2 \pi f / c\) to find the first three natural frequencies of the speaker's vocal tract.

Graph the function. $$ f(x)=\left|\frac{1}{x}-2\right| $$

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