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Using the Definition of the Derivative, Find the Derivative... |
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| Jul1-12, 06:18 PM | #1 |
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Using the Definition of the Derivative, Find the Derivative...
1. The problem statement, all variables and given/known data
Using the definition of the derivative, find the derivative of g(t) = 1 / sqrt(t). 2. Relevant 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. 3. 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)) |
| Jul1-12, 07:22 PM | #2 |
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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] |
| Jul1-12, 07:28 PM | #3 |
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Recognitions:
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Note: [tex]\frac{1}{a}-\frac{1}{b} = \frac{b-a}{ab}[/tex] |
| Jul1-12, 11:21 PM | #4 |
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Using the Definition of the Derivative, Find the Derivative...
Alrighty then. I get it now. Thanks to the both of you.
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| Jul3-12, 12:16 AM | #5 |
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Recognitions:
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No worries.
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| Jul5-12, 01:09 AM | #6 |
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1. The problem statement, all variables and given/known data
I need to find the derivative of t / (t - 1)^2 (not using the definition though). I need to use quotient rule. 2. Relevant 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. 3. The attempt at a solution I already have the solution in an unsimplified form, so there. |
| Jul5-12, 01:26 AM | #7 |
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Mentor
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| Jul5-12, 01:29 AM | #8 |
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| Jul5-12, 01:44 AM | #9 |
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Mentor
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(a - b)(a + b) = a2 - b2
Work with -t2 + 1 a bit to get -t2 + 1 = 1 - t2 = (1 - t)(1 + t) . |
| Jul5-12, 02:00 AM | #10 |
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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.
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| Jul9-12, 12:15 AM | #11 |
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1. The problem statement, all variables and given/known data
Find the derivative of cot(x) / e^t. 2. Relevant 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...x%29+%2F+e%5Et 3. The attempt at a solution Just the function plugged into the quotient rule. |
| Jul9-12, 12:44 AM | #12 |
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Recognitions:
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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. |
| Jul9-12, 01:20 AM | #13 |
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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.
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