Math Problem Statement

Can you use as much detail as possible for how to solve

Find the derivative \(\frac{dv}{dt}\) if \(v = t + \frac{9}{t}\).

Solution

Let’s solve the problem step by step with detailed explanations.

Problem Statement

We are asked to find the derivative dvdt\frac{dv}{dt} where: v=t+9t.v = t + \frac{9}{t}.

Step 1: Understand the given expression

The function vv consists of two terms:

  1. The first term is tt, which is a linear function of tt.
  2. The second term is 9t\frac{9}{t}, which is a rational function.

To find dvdt\frac{dv}{dt}, we need to differentiate both terms with respect to tt.


Step 2: Differentiate tt

The derivative of tt with respect to tt is straightforward: ddt(t)=1.\frac{d}{dt}(t) = 1.


Step 3: Differentiate 9t\frac{9}{t}

For the second term, we use the power rule. Rewrite 9t\frac{9}{t} as: 9t=9t1.\frac{9}{t} = 9t^{-1}.

The power rule states that the derivative of tnt^n is ntn1n \cdot t^{n-1}. Applying this rule to 9t19t^{-1}: ddt(9t1)=9(1)t2=9t2.\frac{d}{dt}(9t^{-1}) = 9 \cdot (-1) \cdot t^{-2} = -\frac{9}{t^2}.


Step 4: Combine the results

Now, sum the derivatives of the two terms: dvdt=ddt(t)+ddt(9t).\frac{dv}{dt} = \frac{d}{dt}(t) + \frac{d}{dt}\left(\frac{9}{t}\right). Substitute the results: dvdt=1+(9t2).\frac{dv}{dt} = 1 + \left(-\frac{9}{t^2}\right).

Simplify: dvdt=19t2.\frac{dv}{dt} = 1 - \frac{9}{t^2}.


Step 5: Verify the answer

From the choices given, the correct answer is: 19t2.\boxed{1 - \frac{9}{t^2}}.


Recap

  1. The derivative of tt is 11.
  2. The derivative of 9t\frac{9}{t} is 9t2-\frac{9}{t^2}.
  3. Adding these results gives dvdt=19t2\frac{dv}{dt} = 1 - \frac{9}{t^2}.

Would you like further clarification or more examples?


5 Related Questions:

  1. What is the derivative of 1t\frac{1}{t}?
  2. How do you find the second derivative of v=t+9tv = t + \frac{9}{t}?
  3. Can you explain the power rule in differentiation?
  4. What is the derivative of a general rational function atn\frac{a}{t^n}?
  5. How would this problem change if the function was v=t9tv = t - \frac{9}{t}?

Tip:

Whenever dealing with terms like 1t\frac{1}{t}, rewrite them as t1t^{-1} to make differentiation easier!

Ask a new question for Free

By Image

Drop file here or Click Here to upload

Math Problem Analysis

Mathematical Concepts

Differentiation
Power Rule
Rational Functions

Formulas

\(\frac{d}{dt}(t) = 1\)
\(\frac{d}{dt}(t^n) = n \cdot t^{n-1}\)
\(\frac{d}{dt}\left(\frac{9}{t}\right) = -\frac{9}{t^2}\)

Theorems

Power Rule of Differentiation

Suitable Grade Level

Grades 11-12 or Introductory Calculus