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Chapter 3: Problem 28

Find the derivative. Assume \(a, b, c, k\) are constants. $$y=3 t^{2}+\frac{12}{\sqrt{t}}-\frac{1}{t^{2}}$$

Short Answer

Expert verified
The derivative is \(y' = 6t - 6t^{-3/2} + 2t^{-3}\).

Step by step solution

01

Understand the Derivative Rule

To find the derivative of a function, we need to know the derivative rules: for \(t^n\), the derivative is \(nt^{n-1}\). Also, the derivative of a constant times a function is the constant times the derivative of the function.
02

Apply the Power Rule to Each Term

For the first term \(3t^2\), apply the power rule: the derivative is \(3 \times 2t^{2-1} = 6t\).
03

Simplify the Second Term

Rewrite \(\frac{12}{\sqrt{t}}\) as \(12t^{-1/2}\). Then, take the derivative using the power rule: \(\frac{d}{dt}(12t^{-1/2}) = 12 \times -\frac{1}{2}t^{-3/2} = -6t^{-3/2}\).
04

Apply the Power Rule to the Third Term

Rewrite \(-\frac{1}{t^2}\) as \(-t^{-2}\). The derivative is \(-2t^{-3}\).
05

Combine the Derivatives

Add up all the derivatives from each term: \(y' = 6t - 6t^{-3/2} + 2t^{-3}\).

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Key Concepts

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

Understanding the Power Rule
The power rule is a cornerstone in calculus when it comes to finding the derivative of polynomial functions. It provides a straightforward method for differentiating expressions of the form \(t^n\). According to the power rule, the derivative of \(t^n\) is \(nt^{n-1}\). This means you simply multiply the exponent by the coefficient and then subtract one from the exponent.

Let's break it down with an example: For the function \(3t^2\), we apply the power rule as follows:
  • Multiply the exponent (2) by the coefficient (3) to get 6.
  • Then, decrease the exponent by one, resulting in \(t^{1}\).
  • The derivative is thus \(6t\).
The power rule simplifies the calculation of derivatives, making it a highly efficient tool for polynomial functions.
Defining the Function Derivative
A function derivative is essentially the rate at which a function's value changes with respect to changes in its input value. In the simplest terms, it's a way of measuring how sensitive a function is to changes in the variable.
  • The derivative provides us with the slope of the tangent line to the curve at any given point.
  • For example, if you have a function representing the position of a car over time, the derivative would represent the car's speed.
In our exercise, we have the function \(y = 3t^2 + \frac{12}{\sqrt{t}} - \frac{1}{t^2}\). To find the derivative, we needed to apply rules such as the power rule to each term. By doing this, we derived \(y' = 6t - 6t^{-3/2} + 2t^{-3}\), which represents the rate of change for the function's various components.
Mastering Calculus Problem-Solving
Calculus problem-solving often feels like piecing together a puzzle. It requires both understanding the concepts and practicing their application. Successful problem-solving in calculus involves several key steps:
  • First, identify the type of function or expression you are dealing with. Is it a polynomial, trigonometric, or exponential?
  • Next, choose appropriate derivative rules, such as the power rule or product rule, based on the function's form.
  • Break down the function into simpler parts if needed. For instance, rewrite fractional or square root expressions in exponential form.
  • Calculate the derivatives for each part and then combine them to find the derivative of the entire function.
  • Simplify your final answer to make it as comprehensible as possible.
With practice, you can gain confidence in your calculus skills, making derivative calculation a more manageable task. Remember, learning calculus is like learning a new language; it takes time and patience to master.

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Most popular questions from this chapter

The world's population \(^{3}\) is about \(f(t)=6.91 e^{0.011 t}\) billion, where \(t\) is time in years since \(2010 .\) Find \(f(0)\) \(f^{\prime}(0), f(10),\) and \(f^{\prime}(10) .\) Using units, interpret your answers in terms of population.

Find the equation of the tangent line to the graph of \(f(x)=\frac{2 x-5}{x+1}\) at the point at which \(x=0.\)

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Show that the relative rate of change of a product \(f g\) is the sum of the relative rates of change of \(f\) and \(g\).
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