Finding the slope using the method of First Principles

In summary: When you rearrange it like that, all the fractional powers cancel out, and you end up with the correct answer.In summary, @ttpp1124's homework statement contains errors in equations and calculations, and does not support the final answer.
  • #1
ttpp1124
110
4
Homework Statement
I solved it, can anyone see if my method is correct?
Relevant Equations
n/a
IMG_3906.jpg
 
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  • #2
ttpp1124 said:
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
@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
Mark44 said:
@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
ttpp1124 said:
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.
 

What is the method of First Principles for finding slope?

The method of First Principles, also known as the method of limits or the method of infinitesimals, is a mathematical approach used to find the slope of a curve at a specific point. It involves taking the limit of the difference quotient as the change in x approaches 0.

Why is the method of First Principles important for finding slope?

The method of First Principles is important because it allows us to find the slope of a curve at any point, even if the function is not a straight line. This method is also the foundation for more advanced concepts in calculus.

How do you use the method of First Principles to find slope?

To use the method of First Principles, you first need to find the difference quotient by subtracting the y-values of two points on the curve and dividing by the difference in their x-values. Then, you take the limit of this difference quotient as the change in x approaches 0, which will give you the slope at that specific point.

Can the method of First Principles be used for any type of curve?

Yes, the method of First Principles can be used for any type of curve, including linear, quadratic, and exponential functions. However, the calculations may become more complex for more complicated curves.

What are the limitations of using the method of First Principles to find slope?

The method of First Principles can be time-consuming and tedious, especially for more complex curves. It also requires a good understanding of limits and may not always yield an exact answer due to the presence of infinitesimal quantities.

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