Recent content by UsernameValid

Well, the problem says h(x). (d/dx)ln(x+(x2-1)1/2*(d/dx)(x+(x2-1)1/2*(d/dx)(x2-1)1/2*(d/dx)(x2-1). 1/(x+(x2-1)1/2 * 1+[x/(x2-1)]1/2 * x/(x2-1)1/2 * 2x Which actually gets me 2x2/(x2-1)

How do I find the derivative of (x)=ln(x+(x^2-1)1/2) The answer is suppose to be 1/(x2-1). But I keep ending up with 2x/(x2-1).

Nevermind, I figured it out.

Where does the normal line to the ellipse x2-xy+y2=3 at the point (-1,1) intersect the ellipse a second time?So I took the derivative of the equation to get: y'=(2x-y)/(x-2y) Then I put (-1,1) into the equation, to get a slope of 1. So, for the normal line I got a slope of -1, which has the...

tan(x-y)=y/(1+x2) When I take the derivatives of both sides, I get: sec2(x-y)(1-y')=[(1+x2)y'-2xy]/(1+x2)2

Equation: eysinx=x+xy I took the derivative of both sides. For the side with eysinx, I used the product rule and chain rule to get: ey*cosx + ey*sinx*y' For the side with x+xy, I used the sum and product rule to get 1+y+xy' So my resulting equation is: ey*cosx + ey*sinx*y'=1+y+xy', which...

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.

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.

I ended up with the derivatives: x2/[(2-x2)(squareroot of 2-x2)] + 1/(squareroot of 2-x2) -x[SUP]2/[(2-x[SUP]2)(squareroot of 2-x[SUP]2)] - 1/(squareroot of 2-x[SUP]2) 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...

Thanks! I understand it now. However, I've run into another problem: https://www.physicsforums.com/showthread.php?p=4119729#post4119729.

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.

Find the derivative of 101-x2. 1.) So, I used power rule to get: (1-x2)(10-x2). 2.) Then I did chain rule and multiplied my previous answer with (d/dx)(-x2). 3.) The answer I get is: (-2x)(1-x2)(10-x2). When I use my calculator, the answer I get is: -20*ln(10)*10-x2. So, where does...