Finding the slope using the method of First Principles

  • Thread starter ttpp1124
  • Start date
  • #1
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4
Homework Statement:
I solved it, can anyone see if my method is correct?
Relevant Equations:
n/a
IMG_3906.jpg
 

Answers and Replies

  • #2
ehild
Homework Helper
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Homework Statement:: I solved it, can anyone see if my method is correct?
Relevant Equations:: n/a

View attachment 261558
It looks correct, but is it "first principle"? Multiply both the numerator and denominator by the sum of the square roots, and take the limit.
 
  • #3
35,222
7,038
@ttpp1124, you ended up with sort of the right answer, but the work shown doesn't support your answer. At the end of your work you have ##2x - 1^{(-1/2)}##. This is technically incorrect. Although you have used parentheses, you put them in the wrong place. Instead, they should be around the expression that's being raised to the power; i.e., like this ##(2x -1)^{-1/2}##. What you wrote would simplify to 2x - 1.

Regarding @ehild's comment about first principles, it doesn't look like you actually evaluated the limit at the bottom of the left half of your work. The part below "Expanding" doesn't make any sense at all -- it looks like you used some sort of product rule. For one thing, that rule doesn't apply here, and for another, you're supposed to find the derivative by first principles; i.e., by using the limit definition of the derivative.

The work at the top of the right half of the page is completely wrong:
##(2x - 1)^{1/2} + (1/2)(2h)(2x - 1)^{(-1/2)}##, and is entirely unrelated to the problem you're doing. If your instruction is even halfway careful, you won't get credit for this work.
 
  • #4
110
4
@ttpp1124, you ended up with sort of the right answer, but the work shown doesn't support your answer. At the end of your work you have ##2x - 1^{(-1/2)}##. This is technically incorrect. Although you have used parentheses, you put them in the wrong place. Instead, they should be around the expression that's being raised to the power; i.e., like this ##(2x -1)^{-1/2}##. What you wrote would simplify to 2x - 1.

Regarding @ehild's comment about first principles, it doesn't look like you actually evaluated the limit at the bottom of the left half of your work. The part below "Expanding" doesn't make any sense at all -- it looks like you used some sort of product rule. For one thing, that rule doesn't apply here, and for another, you're supposed to find the derivative by first principles; i.e., by using the limit definition of the derivative.

The work at the top of the right half of the page is completely wrong:
##(2x - 1)^{1/2} + (1/2)(2h)(2x - 1)^{(-1/2)}##, and is entirely unrelated to the problem you're doing. If your instruction is even halfway careful, you won't get credit for this work.
The end is cut off a bit, my apologies; I think this is better. What do you think?
IMG_3913 2.jpg
 
Last edited by a moderator:
  • #5
35,222
7,038
I think this is better. What do you think?
No, it's not better.
You have a small mistake in line 4 on the left side. In the numerator, you have ##[(2(x + h - 1)^{1/2} - (2x - 1)^{1/2}]##. 2(x + h - 1) is wrong. Also, there are 3 left parens and 2 right parens in the numerator, so that's a mistake. It looks like you caught your error in the 5th line.

The bigger problem is that you apparently know how this should come out, but aren't able to show the work that supports it. In the next to last line, the big ugly fraction is correct, but not the line that follows it.
The reason you're doing what @ehild suggested is to get rid of the fractional powers in the numerator, using the basic idea that ##(x^{1/2} + y^{1/2})(x^{1/2} - y^{1/2}) = x - y##. This is really the formula ##(a + b)(a - b) = a^2 - b^2## in disguise.
 

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