Using the Definition of the Derivative, Find the Derivative

In summary, the conversation discusses finding the derivative of various functions, including g(t) = 1 / sqrt(t), t / (t - 1)^2, and cot(x) / e^t. The quotient rule is used to find the derivative, and factors such as the difference of squares are utilized to simplify the final solution. The conversation also highlights the importance of using the correct variables when finding derivatives.
  • #1
communitycoll
45
0

Homework Statement


Using the definition of the derivative, find the derivative of g(t) = 1 / sqrt(t).

Homework Equations


I was told I could solve it by rationalizing it. I asked a question on Yahoo! Answers and saw someone work it out step by step, but I don't understand any of why they did what they did. If it's painfully obvious, and it's just a matter of knowing how to do some easy algebra please forgive me; it's been a little over a year since my last math class.

The Attempt at a Solution


The Yahoo! Answers person's attempt at the solution that is.

g'(t) = lim[h→0] (g(t+h) − g(t)) / h

g'(t) = lim[h→0] (1/√(t+h) − 1/√(t)) / h

g'(t) = lim[h→0] √(t+h)√(t) * (1/√(t+h) − 1/√(t)) / (h √(t+h)√(t))

I don't understand this ^ step because I don't know where √(t+h)√(t) is coming from, or why it's also being applied to the denominator.

g'(t) = lim[h→0] (√(t) − √(t+h)) / (h √(t+h)√(t))

I don't understand how multiplying the numerator by √(t+h)√(t) got √(t) − √(t+h) either. After this point I think I understand what they're doing.

g'(t) = lim[h→0] (√(t) − √(t+h)) (√(t) + √(t+h)) / (h √(t+h)√(t) (√(t) + √(t+h)))

g'(t) = lim[h→0] (t − (t+h) / (h √(t+h)√(t) (√(t) + √(t+h)))

g'(t) = lim[h→0] −h / (h √(t+h)√(t) (√(t) + √(t+h)))

g'(t) = lim[h→0] −1 / (√(t+h)√(t) (√(t) + √(t+h)))

g'(t) = −1 / (√(t)√(t) (√(t) + √(t)))

g'(t) = −1 / (t * 2√(t))

g'(t) = −1 / (2t^(3/2))
 
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  • #2
As a general rule, you don't want fractions on top of fractions. Your helper multiplied the numerator by something they knew would get rid of the fractions there (the product of the denominators of those fractions). They also had to multiply the denominator by the same quantity so that what they did was a fancy way of multiplying by 1 - i.e. they did nothing, but in a way that left the equation simpler. It's perhaps easier to see typeset:
[tex]
\frac{1/\sqrt{t+h} − 1/\sqrt{t}}{h}=
\left(\frac{\sqrt{t+h}\sqrt{t}}{\sqrt{t+h}\sqrt{t}}\right)\left(\frac{1/\sqrt{t+h} − 1/\sqrt{t}}{h}\right)
[/tex]
 
  • #3
I don't understand this ^ step because I don't know where √(t+h)√(t) is coming from, or why it's also being applied to the denominator.
It's being applied to the numerator because it is also being applied to the denominator ;) the choice for the multiplier was to get rid of the fraction as Ibix said.
I don't understand how multiplying the numerator by √(t+h)√(t) got √(t) − √(t+h) either.
Did you try multiplying it out to see?

Note: [tex]\frac{1}{a}-\frac{1}{b} = \frac{b-a}{ab}[/tex]
 
  • #4
Alrighty then. I get it now. Thanks to the both of you.
 
  • #5
No worries.
 
  • #6

Homework Statement


I need to find the derivative of t / (t - 1)^2 (not using the definition though). I need to use quotient rule.

Homework Equations


I got (-t^2 + 1) / (t - 1)^4, which Wolfram Alpha says can be simplified to -(t+1) / (t-1)^3, which is what people are telling me the answer is. I don't understand how it simplifies to that though, which is what I need to be explained.

The Attempt at a Solution


I already have the solution in an unsimplified form, so there.
 
  • #7
communitycoll said:

Homework Statement


I need to find the derivative of t / (t - 1)^2 (not using the definition though). I need to use quotient rule.

Homework Equations


I got (-t^2 + 1) / (t - 1)^4, which Wolfram Alpha says can be simplified to -(t+1) / (t-1)^3, which is what people are telling me the answer is. I don't understand how it simplifies to that though, which is what I need to be explained.

The Attempt at a Solution


I already have the solution in an unsimplified form, so there.
Do you know how to factor the difference of squares?
 
  • #8
SammyS said:
Do you know how to factor the difference of squares?

I probably used to : D but it's been awhile, as I took precal my junior year and nothing my senior year of HS. Is it just a matter of looking up a Khan Academy video?
 
  • #9
(a - b)(a + b) = a2 - b2

Work with -t2 + 1 a bit to get

-t2 + 1 = 1 - t2 = (1 - t)(1 + t) .
 
  • #10
Okay-dokey. I remember that now. Thank you, it's just that it's been a good long while since I've been in a math class : D I appreciate it.
 
  • #11

Homework Statement


Find the derivative of cot(x) / e^t.

Homework Equations


Using the quotient rule:

-[csc^2(x) + cot(x)] / e^t.

Wolfram Alpha tells me that it's [-e^(-t)][csc^2(x)], not using the quotient rule:

http://www.wolframalpha.com/input/?i=derivative+(cot+x)+/+e^t

The Attempt at a Solution


Just the function plugged into the quotient rule.
 
  • #12
First of all, it's better to start a new thread with a different problem.

Secondly, you have two different variables, t and x. From your answer, I suspect you meant to use a single variable. If you change the t to an x in both Wolfram and your answer, then the answers will match.
 
  • #13
Ah, terribly sorry. Didn't realize that : D Thank you. Also I just figured people wouldn't want me spamming up the forum. I'll start a new thread next time. Muchas gracias.
 

Related to Using the Definition of the Derivative, Find the Derivative

1. What is the definition of the derivative?

The definition of the derivative is the instantaneous rate of change of a function at a specific point. It is the slope of the tangent line at that point and can be calculated using the limit of the difference quotient.

2. How do you find the derivative using the definition?

To find the derivative using the definition, you need to take the limit of the difference quotient as the change in x approaches 0. This will give you the slope of the tangent line at the specific point, which is the value of the derivative.

3. Why is it important to use the definition of the derivative?

Using the definition of the derivative allows us to find the exact rate of change at a specific point, which is crucial in many applications such as physics, economics, and engineering. It also helps us understand the behavior of functions and their graphs.

4. Are there any shortcuts to finding the derivative?

Yes, there are various shortcuts or rules that can be used to find the derivative of specific types of functions, such as the power rule, product rule, quotient rule, and chain rule. These shortcuts can make the process of finding derivatives quicker and easier.

5. How can I check if my answer for the derivative is correct?

You can check your answer by taking the derivative using other methods, such as using the shortcut rules or using software like Wolfram Alpha. You can also graph the original function and its derivative to visually see if they match up.

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