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Chapter 1: Problem 113

Differentiate each function. $$y(t)=5 t(t-1)(2 t+3)$$

Short Answer

Expert verified
y'(t) = 30t^2 + 10t - 15

Step by step solution

01

- Expand the function

First, expand the given function before differentiating. Start by multiplying the factors: \[ y(t) = 5t(t-1)(2t+3) \] Distribute and simplify: \[ = 5t(t(2t+3) - (2t+3)) \] \[ = 5t(2t^2 + 3t - 2t - 3) \] \[ = 5t(2t^2 + t - 3) \] \[ = 10t^3 + 5t^2 - 15t \]
02

- Differentiate term-by-term

Differentiate each term of the expanded function: \[ y(t) = 10t^3 + 5t^2 - 15t \] Apply the power rule \( \frac{d}{dt}[t^n] = nt^{n-1} \): \[ \frac{d}{dt}[10t^3] = 30t^2 \] \[ \frac{d}{dt}[5t^2] = 10t \] \[ \frac{d}{dt}[-15t] = -15 \]
03

- Combine the results

Combine the differentiated terms to form the final answer: \[ y'(t) = 30t^2 + 10t - 15 \]

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Key Concepts

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

calculus
Calculus is a branch of mathematics that explores how things change. It involves two major concepts: differentiation and integration. Differentiation focuses on finding the rate at which something changes. Integration, on the other hand, focuses on finding the total amount accumulated. Differentiation is essential in understanding various physical phenomena, from the speed of a car to the growth rate of a population.
Differentiating a function means finding its derivative. A derivative represents the slope or rate of change of the function at any given point. For instance, if you have a position function, its derivative is the velocity function.
expansion of polynomials
Expanding a polynomial involves multiplying its factors to write it as a sum of terms. This is often a crucial step to simplify complex functions before differentiating them. Consider the given expression: \(y(t) = 5t(t-1)(2t+3)\). The first step is to distribute the terms:
\(y(t) = 5t(2t^2 + t - 3)\).
This process of distributing and simplifying helps break down the polynomial into its basic components.
So, \(5t \times 2t^2\) gives \(10t^3\), \(5t \times t\) gives \(5t^2\), and finally, \(5t \times -3\) gives \(-15t\).
Afterward, you combine these results to get the expanded form: \(y(t) = 10t^3 + 5t^2 - 15t\).
power rule
The Power Rule is a fundamental tool in calculus for differentiating functions. It states that if you have a function in the form \(t^n\), its derivative is \(nt^{n-1}\).
Let's apply the power rule to the function in this exercise, \(y(t) = 10t^3 + 5t^2 - 15t\).
  • For \(10t^3\), the derivative is \(30t^2\).
  • For \(5t^2\), the derivative is \(10t\).
  • For \(-15t\), the derivative is \(-15\).
Combining these, we get the derivative: \(y'(t) = 30t^2 + 10t - 15\).
The power rule simplifies the differentiation process, especially for polynomial functions, making it a powerful tool for calculus students.

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