Derivatives: Composites, normal lines, n-th derivatives and more.

In summary, the conversation is discussing finding the x-intercept of a line perpendicular to a given curve at a specific point, finding the derivative of a composite function, and finding the 98th derivative of a function at a given point. The conversation also includes a discussion on algebraic mistakes and tips for finding the slope of a perpendicular line.
  • #1
StopWatch
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Homework Statement



1. The line perpendicular to the curve y = 2x^3 - x^2 + x - 3 at the point (1, -1) will intersect the x-axis at what point?

2. f(x) = |x^2 - 5| - x, for all x. Let g = f(f(f(x))), find g'(2). I tried just subbing in 2x - 1, the first derivative, to f(2x - 1) and then once more and ended up with 16 somehow, when the answer is -45.

3. If f(x) = ln(2X^2 + x - 1) - ln(x+1) find the 98th derivative at (1/2 + sqrt(2)/2). I know that the derivative simplifies to 2(2x - 1) and the 2nd derivative to -2/(2x-1)^2 and the 3rd to -8(2x-1)/2x-1)^4 but I always have a hard time generalizing these and then getting the answer (especially because the answer is -2^49(97!) and I have no idea how the factorial gets worked in.


Homework Equations



y1 - y0 = m(x1 - x0)



The Attempt at a Solution



Finding the derivative and subbing in x = 1 gives a slope of 5 at the point specified, which means m = -1/5. When solved this gives x = -10, however the correct answer is apparently -4.

This site is a godsend.
 
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  • #2
Hi StopWatch! :smile:
StopWatch said:
1. The line perpendicular to the curve y = 2x^3 - x^2 + x - 3 at the point (1, -1) will intersect the x-axis at what point?

Finding the derivative and subbing in x = 1 gives a slope of 5 at the point specified, which means m = -1/5.

Yes, but how did you then get -10 starting from x = 1, y = -1? :confused:
 
  • #3
StopWatch said:
2. f(x) = |x^2 - 5| - x, for all x. Let g = f(f(f(x))), find g'(2). I tried just subbing in 2x - 1, the first derivative …

But sometimes it's -2x - 1, isn't it? :wink:
3. If f(x) = ln(2X^2 + x - 1) - ln(x+1) find the 98th derivative at (1/2 + sqrt(2)/2). I know that the derivative simplifies to 2(2x - 1) and the 2nd derivative to -2/(2x-1)^2 and the 3rd to -8(2x-1)/2x-1)^4 but I always have a hard time generalizing these and then getting the answer (especially because the answer is -2^49(97!) and I have no idea how the factorial gets worked in.

Hint: what's the nth derivative of 1/x ? :wink:
 
  • #4
Is the nth derivative (n!)x^-n? Does that make sense. Sorry for leaving for so long I had class. I hope someone's around though, the test is tomorrow morning and I have a ton of questions. I wish professors posted solutions to past tests. I redid the algebra and got 10 by the way y (which is zero when the x-axis is intercepted) + 1 = 5(x - 1), so 0 = 5x -2 = 2/5 somehow I really screwed that up, wow.

It can be negative, but we're subbing in when x = 2 so those cases don't matter, or at least that's my logic. My main concern is about how to actually go about plugging into the composite like that.

I do have a new question as well though: The line perpendicular to x^3 - 2x + 1 at (2, 5) will intersect the x-axis at what point? I get (y-5) = 10(x-2) which gives me -15/10 for the x value of the tangent. I might just be really rushed in my thinking right now I'd really appreciate any help at all.
 
  • #5
Anyone around?
 
  • #6
Hi StopWatch! :smile:

(just got up :zzz: …)
StopWatch said:
Is the nth derivative (n!)x^-n?

yeees, but n-1 and times (-1)n :wink:
… 1 = 5(x - 1), so 0 = 5x -2 …

nooo, 5x - 6 :redface:

but anyway it's (x - 1) = -5(1)
The line perpendicular to x^3 - 2x + 1 at (2, 5) will intersect the x-axis at what point? I get (y-5) = 10(x-2) …

no, the slope of the tangent line is 10, so for the perpendicular line you need -10(y-5) = (x-2) :smile:

Anyway, good luck on your test this morning! :smile:
 

1. What are composite derivatives?

Composite derivatives refer to the derivative of a function that is composed of two or more functions. It is calculated using the chain rule, which states that the derivative of a composite function is equal to the product of the derivatives of its component functions.

2. How do I find the normal line of a function?

The normal line of a function is a line that is perpendicular to the tangent line at a given point on the function. To find the normal line, you can use the point-slope form of a line and the derivative of the function at that point. The slope of the normal line is the negative reciprocal of the derivative at that point.

3. What is the n-th derivative of a function?

The n-th derivative of a function is the derivative of the derivative of the function n times. It is denoted by f(n)(x) and can be calculated using the power rule, product rule, quotient rule, and chain rule.

4. How are derivatives used in real-world applications?

Derivatives are used in a variety of fields such as physics, engineering, economics, and finance to analyze and model change. They can be used to calculate rates of change, find maximum and minimum values, and optimize functions.

5. Are there any limitations to using derivatives?

While derivatives are a powerful tool in mathematics, they do have some limitations. They may not exist for certain functions, such as those with sharp corners or discontinuities. Additionally, they cannot always accurately predict the behavior of a function at extreme values or in complex systems.

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