/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 9 \(3-32\) Differentiate the funct... [FREE SOLUTION] | 91Ó°ÊÓ

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\(3-32\) Differentiate the function. \(f(t)=\frac{1}{4}\left(t^{4}+8\right)\)

Short Answer

Expert verified
The derivative is \( f'(t) = t^3 \).

Step by step solution

01

Recognize the Function Structure

The given function is \( f(t) = \frac{1}{4}(t^4 + 8) \). It is a polynomial inside a constant multiplier, making it relatively straightforward to differentiate.
02

Apply the Constant Multiple Rule

According to the constant multiple rule, if a function is multiplied by a constant, the derivative of the function is the constant times the derivative of the function itself. Here, the constant is \( \frac{1}{4} \), so we place this outside the differentiation operation: \( f'(t) = \frac{1}{4} \cdot \frac{d}{dt}(t^4 + 8) \).
03

Differentiate the Polynomial

We differentiate the polynomial \( t^4 + 8 \). The derivative of \( t^4 \) is \( 4t^3 \) using the power rule, \( nx^{n-1} \). The derivative of a constant (8) is 0. Therefore, \( \frac{d}{dt}(t^4 + 8) = 4t^3 + 0 = 4t^3 \).
04

Simplify the Expression

Plug the derivative \( 4t^3 \) back into the expression with the constant multiplier \( \frac{1}{4} \): \( f'(t) = \frac{1}{4} \cdot 4t^3 = t^3 \).

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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 fundamental concept in calculus that makes differentiating power functions straightforward and efficient.
It is especially useful when dealing with polynomial expressions.
According to the power rule, the derivative of a function of the form \( t^n \) is given by \( nt^{n-1} \). This means that you simply multiply the exponent by the variable and reduce the power by one.
For example, if you need to differentiate \( t^4 \), you apply the power rule by multiplying the exponent 4 by \( t^{(4-1)} \), resulting in \( 4t^3 \).
This rule helps in rapidly handling polynomial terms, simplifying complex expressions into manageable pieces.
So, whenever you differentiate terms like \( t^2 \) or \( t^6 \), remember the power rule for an efficient solution.
Polynomial Differentiation Simplified
Polynomial differentiation involves the application of basic differentiation rules to polynomial expressions.
A polynomial function is usually presented as a sum of terms such as \( t^4 + 8 \), where you differentiate each term individually.
Here’s how it works: for a term like \( t^4 \), you apply the power rule to find its derivative, which is \( 4t^3 \).
For constant terms like 8, the derivative is always 0 because constants do not change and thus have no rate of change.
The crucial point is to address each term one at a time. This process is systematic, ensuring you do not miss out on any portion of the expression.
Always remember that differentiation of polynomials is just about using the right rule for each part, which makes the overall task much easier.
Applying the Constant Multiple Rule
The constant multiple rule is another essential tool in differentiation which simplifies calculations when a constant is involved.
This rule states that if a function \( f(t) \) is multiplied by a constant, say \( c \), then the derivative of the function is simply that constant multiplied by the derivative of \( f(t) \).
Consider the function \( f(t) = \frac{1}{4}(t^4 + 8) \).
Here, the constant is \( \frac{1}{4} \). According to the constant multiple rule, you take this constant out and multiply it with the derivative of the polynomial \( t^4 + 8 \).
So you first differentiate \( t^4 + 8 \), which gives you \( 4t^3 \) and then multiply it by \( \frac{1}{4} \) resulting in \( f'(t) = t^3 \).
This rule is particularly useful because it saves time and reduces errors by allowing you to handle constants efficiently right at the start of the differentiation process.

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