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Chapter 2: Problem 11

Find the derivative of the function. $$ y=\left(t+\frac{2}{t}\right)^{6} $$

Short Answer

Expert verified
The derivative of the function \(y=\left(t+\frac{2}{t}\right)^{6}\) is: \(\frac{dy}{dt} = 6\left(t + \frac{2}{t}\right)^5 \cdot \left(1 - \frac{4}{t^2}\right)\)

Step by step solution

01

First, we need to recognize the given function as a composition of two functions: \(y = g(u)\) and \(u = f(t)\). Let's assign them as follows: - For the outer function, let \(g(u) = u^6\). - For the inner function, let \(u = f(t) = t + \frac{2}{t}\). **Step 2: Apply the chain rule**

Now, we'll apply the chain rule and the power rule to find the derivative of \(y\) with respect to \(t\). Using the chain rule: \(\frac{dy}{dt} = \frac{dy}{du} \cdot \frac{du}{dt}\) To find \(\frac{dy}{du}\), apply the power rule on \(y = g(u) = u^6\): \(\frac{dy}{du} = 6u^5\) Next, we'll find \(\frac{du}{dt}\) by differentiating \(u = f(t) = t + \frac{2}{t}\) with respect to \(t\). **Step 3: Differentiate the inner function**
02

Using the power rule and the fact that the derivative of \(\frac{1}{t}\) is \(-\frac{1}{t^2}\), we have: - Derivative of \(t\) is 1. - Derivative of \(\frac{2}{t}\) is \(2 \cdot (-\frac{1}{t^2}) = -\frac{4}{t^2}\). Therefore, the derivative of \(u = f(t) = t + \frac{2}{t}\) with respect to \(t\) is: \(\frac{du}{dt} = 1 - \frac{4}{t^2}\) **Step 4: Find the derivative of y with respect to t**

Now we have all the required pieces to compute the derivative of the given function. Use the chain rule formula and substitute the values we found in the previous steps: \(\frac{dy}{dt} = \frac{dy}{du} \cdot \frac{du}{dt} = 6u^5 \cdot \left(1 - \frac{4}{t^2}\right)\) As a final step, we'll substitute \(u = f(t) = t + \frac{2}{t}\) back into the formula to get the derivative of y with respect to t in terms of t: \(\frac{dy}{dt} = 6\left(t + \frac{2}{t}\right)^5 \cdot \left(1 - \frac{4}{t^2}\right)\) So, the derivative of the given function is: \(\frac{dy}{dt} = 6\left(t + \frac{2}{t}\right)^5 \cdot \left(1 - \frac{4}{t^2}\right)\)

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Key Concepts

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

Derivative of a Function
Understanding the derivative of a function is essential in calculus, as it represents the rate at which a function's output changes relative to its input. In simpler terms, the derivative measures how quickly the y-value of a function rises or falls as the x-value moves. It's often thought of as finding the slope of the function at any given point.

For instance, in physics, the derivative of the position with respect to time gives you the velocity, which is the rate of change of position. Similarly, the derivative of velocity with regard to time will give you the acceleration, which is the rate of change of velocity. Calculus offers a toolbox of techniques to compute derivatives for different kinds of functions.
Power Rule Differentiation
One of the key tools in calculus is the power rule, which simplifies the task of finding the derivative of a function where the variable has an exponent, often referred to as a power. The rule states that if you have a function of the form
\(f(x) = x^n\)
, where \(n\) is any real number, the derivative of the function with respect to \(x\) is
\(f'(x) = nx^{n-1}\)
.

This makes computation straightforward for polynomial functions like \(x^6\), as seen in our example, and it's a building block for understanding more complex rules of differentiation.
Composition of Functions
A composition of functions occurs when one function is nested inside another. We express this as \(g(f(x))\), where \(f(x)\) is the inner function and \(g(x)\) is the outer function. For any given x value, you first apply the inner function f, and then you use its output as the input for the outer function g.

In our exercise, \(f(t) = t + \frac{2}{t}\) is nested within \(g(u) = u^6\), creating a composite function \(g(f(t))\). Deconstructing complex functions like these into simpler parts can be key to finding their derivatives.
Chain Rule Application
The chain rule is a principle in calculus used precisely for finding the derivative of composite functions. It bridges the gap when you're not simply dealing with a power or a multiple but rather a function within a function. According to the chain rule, the derivative of a composite function \(g(f(x))\) is the derivative of the outer function evaluated at the inner function, times the derivative of the inner function:
\(\frac{dg}{dx} = \frac{dg}{df} \cdot \frac{df}{dx}\)
.

By using the chain rule in conjunction with the power rule, you can break down complicated derivative problems, just like we did in the given exercise, finding the derivative step by step and ensuring a clear understanding of the function's behavior at any point.

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Most popular questions from this chapter

Let \(g\) denote the inverse of the function \(f\). (a) Show that the point \((a, b)\) lies on the graph of \(f .\) (b) Find \(g^{\prime}(b)\) $$ f(x)=\frac{x+1}{2 x-1} ; \quad(1,2) $$

Find an equation of the tangent line to the graph of the function at the indicated point. Graph the function and the tangent line in the same viewing window. $$ f(x)=x \sin ^{-1} x ; \quad P\left(\frac{1}{2}, \frac{\pi}{12}\right) $$

A projectile is fired from a cannon located on a horizontal plane. If we think of the cannon as being located at the origin \(O\) of an \(x y\) -coordinate system, then the path of the projectile is $$y=\sqrt{3} x-\frac{x^{2}}{400}$$ where \(x\) and \(y\) are measured in feet. a. Find the value of \(\theta\) (the angle of elevation of the gun). b. At what point on the trajectory is the projectile traveling parallel to the ground? c. What is the maximum height attained by the projectile? d. What is the range of the projectile (the distance \(O A\) along the \(x\) -axis)? e. At what angle with respect to the \(x\) -axis does the projectile hit the ground?

A 20 -ft ladder leaning against a wall begins to slide. How fast is the angle between the ladder and the wall changing at the instant of time when the bottom of the ladder is \(12 \mathrm{ft}\) from the wall and sliding away from the wall at the rate of \(5 \mathrm{ft} / \mathrm{sec} ?\)

Suppose that \(u\) is a differentiable function of \(x\) and \(f(x)=|u| .\) Show that $$ f^{\prime}(x)=\frac{u^{\prime} u}{|u|} \quad u \neq 0 $$ Hint: \(|u|=\sqrt{u^{2}}\).
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