Chapter 3: Problem 24
Find the derivative of the following functions. \(s(t)=4 \sqrt{t}-\frac{1}{4} t^{4}+t+1\)
Short Answer
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Question: Find the derivative of the function \(s(t) = 4\sqrt{t} -\frac{1}{4}t^4 + t + 1\).
Answer: The derivative of the function is \(s'(t) = 2t^{-\frac{1}{2}} - t^3 + 1\).
Step by step solution
01
Identify the terms and rules of differentiation
The function consists of four terms: \(4\sqrt{t}\), \(-\frac{1}{4}t^4\), \(t\), and \(1\). We'll apply the appropriate rules of differentiation (power rule, constant rule) to each of these terms.
02
Differentiate the first term (\(4\sqrt{t}\))
To differentiate the first term, first rewrite it using a fractional exponent: \(4\sqrt{t} = 4t^{\frac{1}{2}}\). Now apply the power rule:
\(\frac{d}{dt}(4t^{\frac{1}{2}}) = 4 \cdot \frac{1}{2}t^{\frac{1}{2}-1} = 2t^{-\frac{1}{2}}.\)
03
Differentiate the second term (\(-\frac{1}{4}t^4\))
Apply the power rule:
\(\frac{d}{dt}(-\frac{1}{4}t^4) = -\frac{1}{4} \cdot 4t^{4-1} = -t^3.\)
04
Differentiate the third term (\(t\))
Apply the power rule:
\(\frac{d}{dt}(t) = 1t^{1-1} = 1.\)
05
Differentiate the fourth term (\(1\))
This term is a constant, and the derivative of a constant is always 0:
\(\frac{d}{dt}(1) = 0.\)
06
Combine the derivatives of the individual terms
To find the derivative of the entire function, sum the derivatives of the individual terms:
\(\frac{d}{dt}(s(t)) = 2t^{-\frac{1}{2}} - t^3 + 1 + 0 = 2t^{-\frac{1}{2}} - t^3 + 1\)
So, the derivative of the function is:
\(s'(t) = 2t^{-\frac{1}{2}} - t^3 + 1\).
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Power Rule
The power rule is a fundamental tool in calculus that makes finding derivatives straightforward. Imagine you have a function, and each term of this function is of the form \( ax^n \), where \( a \) is a constant, and \( n \) is an exponent. The power rule states that the derivative of this term is obtained by multiplying the exponent \( n \) by the coefficient \( a \), followed by subtracting one from the exponent.
Here's how it works step-by-step:
For example, in our exercise, terms like \(-\frac{1}{4}t^4\) and \( t \) use the power rule:
Here's how it works step-by-step:
- Identify the coefficient (\( a \)) and the exponent (\( n \)).
- Multiply the coefficient by the exponent.
- Subtract one from the exponent.
For example, in our exercise, terms like \(-\frac{1}{4}t^4\) and \( t \) use the power rule:
- For \(-\frac{1}{4}t^4\), the exponent \(4\) is multiplied by \(-\frac{1}{4}\), resulting in \(-1\), and the exponent is reduced to \(3\).
- The derivative of \(t\) or \(t^1\) is simply 1 because the exponent 1 multiplied by the coefficient 1 remains as 1, and the exponent reduces to 0.
Constant Rule
In calculus, the constant rule simplifies the differentiation process greatly when dealing with constants. A constant is any real number, like \(1\) in our exercise, which doesn't change regardless of the value of the variable. The constant rule is incredibly straightforward: it states that the derivative of any constant term is always zero!
This is because constants do not have a rate of change with respect to the variable. They are fixed values and therefore their "slope" is zero.
Consider the constant 1 in the function \( s(t) = 4\sqrt{t} - \frac{1}{4} t^4 + t + 1 \). Its derivative is \( \frac{d}{dt}(1) = 0 \).
Whenever you see a standalone number in a function, remember that its contribution to the derivative is zero. This simplifies many problems significantly.
This is because constants do not have a rate of change with respect to the variable. They are fixed values and therefore their "slope" is zero.
Consider the constant 1 in the function \( s(t) = 4\sqrt{t} - \frac{1}{4} t^4 + t + 1 \). Its derivative is \( \frac{d}{dt}(1) = 0 \).
Whenever you see a standalone number in a function, remember that its contribution to the derivative is zero. This simplifies many problems significantly.
Fractional Exponents
Fractional exponents are another way of expressing roots. This is a key concept in calculus and is especially useful for differentiation. Instead of writing \( \sqrt{t} \), you can express this as \( t^{\frac{1}{2}} \). This reinterpretation is crucial for using the power rule effectively, as you need the expression in the form \( x^n \) to differentiate.
Changing a root to a fractional exponent allows the same methods of differentiation to be applied across different types of terms.
Consider \( 4\sqrt{t} \) in the initial function. The square root was rewritten as \( 4t^{\frac{1}{2}} \). By doing this, you can apply the power rule to find the derivative easily, calculated as \( 2t^{-\frac{1}{2}} \).
The usage of fractional exponents bridges the gap between polynomials and roots, enabling a unified approach to solving more complex calculus problems. This makes calculus much more manageable even when roots are involved in the function.
Changing a root to a fractional exponent allows the same methods of differentiation to be applied across different types of terms.
Consider \( 4\sqrt{t} \) in the initial function. The square root was rewritten as \( 4t^{\frac{1}{2}} \). By doing this, you can apply the power rule to find the derivative easily, calculated as \( 2t^{-\frac{1}{2}} \).
The usage of fractional exponents bridges the gap between polynomials and roots, enabling a unified approach to solving more complex calculus problems. This makes calculus much more manageable even when roots are involved in the function.