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Chapter 1: Problem 28

Differentiate each function. \(f(t)=\frac{t}{5+2 t}-2 t^{4}\)

Short Answer

Expert verified
\( f'(t) = \frac{5}{(5+2t)^2} - 8t^3 \)

Step by step solution

01

Identify the Functions to Differentiate

The function given is a combination of two parts: \( \frac{t}{5+2t} \) and \( -2t^4 \). We need to differentiate each of these separately.
02

Differentiate the Quotient Function

To differentiate \( \frac{t}{5+2t} \), we apply the quotient rule, which states that \( \left(\frac{u}{v}\right)' = \frac{u'v - uv'}{v^2} \), where \( u = t \) and \( v = 5 + 2t \). First, find \( u' = 1 \) and \( v' = 2 \). Plug these into the quotient rule:\[ f'(t) = \frac{(1)(5+2t) - (t)(2)}{(5+2t)^2} \]Simplify the numerator:\[ f'(t) = \frac{5 + 2t - 2t}{(5+2t)^2} = \frac{5}{(5+2t)^2} \].
03

Differentiate the Polynomial Function

Now differentiate \( -2t^4 \) using the power rule. The power rule states that \( (t^n)' = n t^{n-1} \), so:\[ \frac{d}{dt}(-2t^4) = -2 \times 4t^{4-1} = -8t^3 \].
04

Combine the Derivatives

Add the derivatives from Step 2 and Step 3 to find \( f'(t) \):\[ f'(t) = \frac{5}{(5+2t)^2} - 8t^3 \].

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Key Concepts

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

Quotient Rule
In calculus, the quotient rule is essential when you need to differentiate a function that is a fraction and involves two functions in the form \( \frac{u}{v} \). This rule makes it possible to find the derivative of such a quotient by using the derivatives of the numerator and denominator.The formula for the quotient rule is:\[ \left(\frac{u}{v}\right)' = \frac{u'v - uv'}{v^2} \]Here's how it works:
  • Let \( u \) be the numerator, and \( v \) be the denominator.
  • Find the derivatives \( u' \) of \( u \) and \( v' \) of \( v \).
  • Plug these into the formula to find the derivative of the quotient.
For the function \( \frac{t}{5+2t} \), we identify \( u = t \) and \( v = 5+2t \). Thus:
  • \( u' = 1 \)
  • \( v' = 2 \)
Using the quotient rule:\[ f'(t) = \frac{(1)(5+2t) - (t)(2)}{(5+2t)^2} = \frac{5}{(5+2t)^2} \]This process simplifies the differentiation of fractions into manageable steps.
Polynomial Differentiation
Polynomial differentiation is a straightforward part of calculus. When faced with a polynomial function such as \( -2t^4 \), the goal is to find its derivative. Polynomial functions are those in which variables have constant exponents and coefficients. They are expressed as sums of terms of the form \( a_n t^n \), where \( a_n \) is a constant coefficient and \( n \) is a non-negative integer.To differentiate polynomials:
  • Apply the power rule to each term individually.
  • Add the derivatives of each term to get the derivative of the polynomial.
For the given function \(-2t^4\):
  • Apply the power rule: \((t^n)' = n t^{n-1}\).
  • Thus, \( \frac{d}{dt}(-2t^4) = -2 \times 4t^{4-1} = -8t^3 \).
By following these steps, finding the derivative becomes a methodical task, allowing you to see how each term contributes to the overall rate of change.
Power Rule
The power rule is a fundamental tool in calculus that simplifies the differentiation process. It applies to functions where the variable is raised to a power, such as \( t^4 \). The rule is written as:\[ (t^n)' = n t^{n-1} \]This straightforward rule is particularly useful for polynomial differentiation.To utilize this rule effectively:
  • Identify the power \( n \) of the variable.
  • Multiply by \( n \) and reduce the exponent by one.
For example, applying the power rule to \( -2t^4 \):
  • \( n = 4 \) implies that the derivative is \(-2 \cdot 4t^{4-1} = -8t^3 \).
The beauty of the power rule is its simplicity and utility, making differentiation of power terms quicker and more intuitive. Integrating this rule into calculus practice helps gain agility in solving a variety of calculus problems.

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Most popular questions from this chapter

Let \(f\) and \(g\) be differentiable over an open interval containing \(x=a\). If $$ \lim _{x \rightarrow a}\left(\frac{f(x)}{g(x)}\right)=\frac{0}{0} \quad \text { or } \quad \lim _{x \rightarrow a}\left(\frac{f(x)}{g(x)}\right)=\frac{\pm \infty}{\pm \infty} $$ and if \(\lim _{x \rightarrow a}\left(\frac{f^{\prime}(x)}{g^{\prime}(x)}\right)\) exists, then $$ \lim _{x \rightarrow a}\left(\frac{f(x)}{g(x)}\right)=\lim _{x \rightarrow a}\left(\frac{f^{\prime}(x)}{g^{\prime}(x)}\right) $$ The forms \(0 / 0\) and \(\pm \infty / \pm \infty\) are said to be indeterminate. In such cases, the limit may exist, and l'H么pital's Rule offers a way to find the limit using differentiation. For example, in Example 1 of Section \(1.1,\) we showed that $$ \lim _{x \rightarrow 1}\left(\frac{x^{2}-1}{x-1}\right)=2 $$ Since, for \(x=1,\) we have \(\left(x^{2}-1\right)(x-1)=0 / 0,\) we differentiate the numerator and denominator separately, and reevaluate the limit: $$ \lim _{x \rightarrow 1}\left(\frac{x^{2}-1}{x-1}\right)=\lim _{x \rightarrow 1}\left(\frac{2 x}{1}\right)=2 $$ Use this method to find the following limits. Be sure to check that the initial substitution results in an indeterminate form. $$ \lim _{x \rightarrow-\infty}\left(\frac{3 x^{3}+x+11}{6 x^{3}+x+2}\right) $$

For each function, find the points on the graph at which the tangent line is horizontal. If none exist, state that fact. $$ y=-x^{3}+1 $$

For each function, find the points on the graph at which the tangent line is horizontal. If none exist, state that fact. $$ y=3 x^{2}-5 x+4 $$

Suppose Fast Trends determines that the revenue, in dollars, from the sale of \(x\) iPod holders is given by $$ R(x)=-0.001 x^{2}+150 x $$ Find \(\frac{R(305)-R(300)}{305-300},\) and interpret the significance of this result to the company.

For each function, find the points on the graph at which the tangent line is horizontal. If none exist, state that fact. $$ y=-\frac{1}{3} x^{3}+6 x^{2}-11 x-50 $$
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