How to find the derivative with an absolute value?

[UsernameValid]

I need to find the derivative of y=| x | / squareroot of 2-x².

We never learned how to find the derivative with an absolute value, so I have absolutely no idea how to do this problem and I can't find an example in the textbook.

 

[SammyS]

I need to find the derivative of y=| x | / squareroot of 2-x².

We never learned how to find the derivative with an absolute value, so I have absolutely no idea how to do this problem and I can't find an example in the textbook.

Express this as a piecewise defined function.

 

[Ray Vickson]

I need to find the derivative of y=| x | / squareroot of 2-x².

We never learned how to find the derivative with an absolute value, so I have absolutely no idea how to do this problem and I can't find an example in the textbook.

Look at the two cases x > 0 and x < 0.

RGV

 

[UsernameValid]

I ended up with the derivatives:

x²/[(2-x²)(squareroot of 2-x<sup>2)] + 1/(squareroot of 2-x2)

-x2/[(2-x2)(squareroot of 2-x2)] - 1/(squareroot of 2-x2)


Now, I need to find the tangent line to the curve at (1,1). I'm following an example from the textbook, but I keep ending up with y=2x-1, which passes through the original curve twice. Here's what I did:

1.) I substituted 1 for x in the derivative and got y=2, which I used as the slope.
2.) Then I used slope intercept form to solve for b. I had x=1, y=1, and m=2.
3.) So, I end up with the equation y=2x-1.</sup>

 

[Levi Tate]

Where do you go to school? I'm taking intro to analysis and I've never taken the derivative of the absolute value. I wish I did, your school seems a bit tough. What book do you use? Perhaps I can download it and help you more.

 

[UsernameValid]

Where do you go to school? I'm taking intro to analysis and I've never taken the derivative of the absolute value. I wish I did, your school seems a bit tough. What book do you use? Perhaps I can download it and help you more.

No, you don't want to go to my school. It sucks and I'm transferring next year. And my textbook is Calculus 7E Early Transcendentals by James Stewart. The question is on Chapter 3.4 #56. And, apparently, it can be downloaded online. I wish I had known that before buying my book.

 

[Levi Tate]

Neat i have that book, let me check it out. That's a solid calculus book there. My calculus teacher uses it, he likes it a lot.

 

[HallsofIvy]

I ended up with the derivatives:

x²/[(2-x²)(squareroot of 2-x<sup>2)] + 1/(squareroot of 2-x2)

-x2/[(2-x2)(squareroot of 2-x2)] - 1/(squareroot of 2-x2)</sup>

I assume these are for different values of x. You say what the values of x are. That is an important part of the answer.

Now, I need to find the tangent line to the curve at (1,1).

Since x is positive in small neighborhoods of 1, You can simply ignore the absolute value.

I'm following an example from the textbook, but I keep ending up with y=2x-1, which passes through the original curve twice. Here's what I did:

1.) I substituted 1 for x in the derivative and got y=2, which I used as the slope.

You mean you got y'= 2 at x= 1.

2.) Then I used slope intercept form to solve for b. I had x=1, y=1, and m=2.
3.) So, I end up with the equation y=2x-1.

 

[Levi Tate]

You've got to check out the graph of the function. The absolute value doesn't matter because if you check out the graph, and probably because it's a function and you're trying to find a tangent at any point x, you don't need the absolute value, so let me try to solve that problem here...

I think it's y = 2x +1

Just check your point algebra and then graph both functions on your calculator to check.

 

[Levi Tate]

^your algebra, I was tying point slope form and I'm falling asleep here.

 

[Levi Tate]

^ trying to type point slope form, sorry.

 

[UsernameValid]

Well, I feel dumb. I was graphing (x / squareroot of 2-x2) and (-x / squareroot of 2-x2) separately. I was getting two s-shaped graphs. I could have been finished with this question a long time ago. Well, thanks for your help.

 

[Ray Vickson]

You've got to check out the graph of the function. The absolute value doesn't matter because if you check out the graph, and probably because it's a function and you're trying to find a tangent at any point x, you don't need the absolute value, so let me try to solve that problem here...

I think it's y = 2x +1

Just check your point algebra and then graph both functions on your calculator to check.

You definitely DO need the absolute value. The two functions f1(x) = |x|/sqrt(2-x^2) and f2(x) = x/sqrt(2-x^2) have very different graphs (although the two graphs coincide in the region x > 0). Furthermore, f2(x) is continuously differentiable in the region (-1/sqrt(2) < x < 1/sqrt(2)), while f1(x) is not (it does not have a derivative at x = 0).

RGV

 

[Levi Tate]

I don't follow your reasoning, I'm definitely not saying I'm right, but if it's a function, then at each x value there is a unique output, so on it's graph, at x=1 it has one y coordinate.

Where my reasoning could certainly be flawed is in this, I assumed that the positive region of the absolute value corresponded to positive x values. If this is true, and I do not know, then what I said seems right to me.

 

[Levi Tate]

Wait yeah you are taking this out of the context of the problem, the problem in the book asks you to simply find the tangent line at (1,1), that's all. As for the theory, it's beyond my understanding and the question he or she is asking.

 

[Ray Vickson]

Wait yeah you are taking this out of the context of the problem, the problem in the book asks you to simply find the tangent line at (1,1), that's all. As for the theory, it's beyond my understanding and the question he or she is asking.

Well, he/she first asked how to take the derivative of |x|/sqrt(2-x^2). In a LATER posting, he/she said "Now, I need to find the tangent line to the curve at (1,1)". Of course the tangent line at x = 1 only needs the derivative in the region {x > 0}, but that is not what the original problem said. Maybe the OP did not actually post the precise problem, but I can only go by what is written.

RGV

 

[Levi Tate]

Right, I looked in the textbook, it just says find the tangent line of the original function at (1,1), it's an introductory calculus problem. I don't even understand your analysis of it, but I respect it.

 

[Ray Vickson]

Right, I looked in the textbook, it just says find the tangent line of the original function at (1,1), it's an introductory calculus problem. I don't even understand your analysis of it, but I respect it.

I did not analyse anything; I just suggested breaking up the problem into two parts (x > 0 or x < 0). After all, |x| is defined differently in those two parts, so it makes sense to do that.

RGV