Problem 11 - Single Variable Calculus

Question

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

Step-by-Step Solution

Answer

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 1: 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**

Step 2: 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: \(\frac{dy}{dt} = 6\left(t + \frac{2}{t}\right)^5 \cdot \left(1 - \frac{4}{t^2}\right)\)

Key Concepts

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.