Recent content by KFSKSS

Okay, thanks Mark44. I typed it wrong twice. I meant what you typed. Now, HOW CAN I SOLVE THE PROBLEM?

Sorry I must have written it wrong. a)=(-4x^3)/2y+1. a) and b) are both correct since the book gives that answer too.

Hello. My problem is as follows: Suppose x^4+y^2+y-3=0. a) Compute dy/dx by implicit differentiation. b) What is dy/dx when x=1 and y=1? c) Solve for y in terms of x (by the quadratic formula) and compute dy/dx directly. Compare with your answer in part a). I solved a) and b). a)=-4x^3/2y+1, and...

As epenguin said "we all have our blind moments".

I like this book I'm learning relatively fast with it. Only sometimes the things in it are "not written the way I would understand". Anyway thank you all. And thanks for your patience.

So for example, the linear approximation of (4.62)^3=x=5; Δx=4.62-5=-0.38; (5)^3+3(5)^2·(-0.38)=125+75(-0.38)=125-28.5=96.5. Is this correct?

To get x I choose a whole number close to value on the function and subtract; 2.94, x=2.94-3=-0.06. Is this right?

Where am I failing? 2.94; x=3, Δx=2.94-3=-0.06. But (3)^4+4(3)^3·(-0.06)=11.34??

So x must be a number close to 2.94, which is three. In case 1.2, x=1 right? What about Δx? How do I choose its value?

Scottdave, f=x^4, df/dx= 4x^3.

Dick how did you do it?

What I have been doing until now was this: (2.94)^4=(x+Δx)^4=x^4+4x^3Δx+6x^2Δx^2+4xΔx^3+Δx^4; d/dx(x^4+4x^3Δx+6x^2Δx^2+4xΔx^3+Δx^4)=4x^3+12x^2Δx+12xΔx^2+4Δx^3; now if x=2 and Δx=0.94=4(2)^3+12(2)^2(0.94)+12(2)(0.94)^2+4(0.94)^3=101.648736. Honestly I think the book has an awful explanation of...

Dick, I tried many things and sometimes I get a number under the result or over it. Scottdave, f ' is the derivative of the function f. f '=df/dy. I have no idea how it can give 74.52. Thanks for answering anyway.

Hello. My problem is that I began with Linear Approximation and I'm terribly stuck. I have problems understanding its very concept and with calculations. (It may sound stupid but I'm autodidact and I'm studying mathematics in english (not my mother tongue) and sometimes it gets hard). It would...