derivative of sqrt(x+y)-sqrt(x-y)=1

[AndyHoladay]

In the second method, instead of solving for y it seems like it would be easier to short-circuit at the step y^2 = x - 1/4, do implicit differentiation => 2y dy/dx = 1; immediately giving dy/dx = 1/(2y)

[ChaosPod]

The 2nd way is better as there is less manipulation. Also, when you get to the stage of
x = y^2 + 0.25 you can do
dx/dy = 2y and then
dy/dx = 1/2y

[Sid-ixqr]

7:00 People became happy because the denominator was rationalised!

[szymon]

in my opinion nicer is 2nd but with one diffrence :
when y^2=x-1/4 dont square both sides just diffrentiate and u get dy/dx*2y=1 divide by 2y and you have dy/dx=1/2y ;)

[amalkrishnanri]

Can't we differentiate both sides after reaching y^2 = x - 1/4 ?
Then it will directly lead to
2y dy = dx
dy / dx = 1/2y

[jameswilson]

Nice choice in problem and great video as always. Too bad you didn't realize the minus sign is not part of the solution. When you had 2y - 1 = 2sqrt(x-y), you could have noted this implies y > 1/2. Then it's easy to tell x > y from the original equation.

[AndyWudookey]

Multiply by (sqrt(x+y)+sqrt(x-y)) on both sides and then add the equations to get 2sqrt(x+y) = 2y+1. Square both sides to get 4(x+y) = (2y+1)^2. Implicit: 4 + 4y' = (8y+4)y', so 8yy' = 4, y' = 1/2y

[angelmendez-rivera]

The final answer stated that d(y)/dx = 1/2y. It is trivial then to see that y = x^(1/2). We can multiply by y, letting y*d(y)/dx = 1/2. Then we can integrate, such that (d/dx)^(-1)(y*(d/dx)(y)) = y^2 - (d/dx)^(-1)(1/2) = y^2 - x/2 = x/2 + C, so y^2 = x + C. Now, y = 0 cannot happen, because y started in the denominator, so x + C > 0. Also, y^2 = (-y)^2, so y = +/- (x - C)^(1/2), x > C.

[knotwilg]

If you (a) square the equation, (b) isolate the remaining root part and square again, you get y² = x - 1/4, so that (c) with implicit derivation y'=1/2y

(a) (V(x+y) - V(x-y))² = x+y + x - y + 2V(x²-y²) = 2x + 2V(x²-y²) == 1
(b) 2V(x²-y²) = 1 - 2x => 4(x²-y²) = 4x² - 4x +1 => -4y² = -4x+1 => y² = x - 1/4
(c) 2yy'=1 and y'=1/2y

This method avoids any fractional arithmetic and postpones the chain rule to the very end.

[BSnicks]

At 8:55 I would just differentiate 4y^2+1 = 4x which will give 8y = 4dx/dy

[ChefSalad]

+blackpenredpen you have made a mistake! The ± sign should just be a + sign (or omitted). To see why, look at your equation y = ±√(x−1/4). Let x = 5/4 and you get y = ±1. Put those two points back into the original equation and you find that only (5/4, +1) works. (5/4, −1) does not. This means that y = +√(x−1/4) only, no ±. Always check a special case when a ± appears in your answer. Quite often only one side will work and you can replace the ± with + or a −.

[adb]

If you were using this in some real-life situation (engineering, physics, economics) the 2nd method would be infinitely more useful because you obtain both the dependent variable y (the function) and its derivative in function of the independent variable only. Having the y and the x mixed up, or having dy/dx in function of y only, is not very useful in practice (but it can be solved, if you have dy/dx=1/2y then 2y.dy = dx then integrating y^2 = x+c then y = +/- sqrt (x+c) and then you would need to put some values of x and go to the original equation to find out if you have to go with the + or the - and that c=1).

[PlutoTheSecond]

I prefer the second method because it helps you find the original function as well as its derivative. The first method gives you a differential equation, which is easily solvable:
2y*dy/dx=1
S 2y dy=S dx
y^2=x+C
y=+sqrt(x+C)
But then you have to use the second method anyway to determine the value of C.

[moskthinks]

You can actually find the value of y using the derivative and initial value.
y'=1/2y.
2ydy = dx.
y^2=x+c
Now we need to evaluate c, which happens to equal -1/4.
y=(-/+)sqrt(x-1/4)
: )

[sanjeev]

Sir please solve this
Frank and sofia investigate two positive integer A and B as follows:
1)frank picks a positive factor a of A
2) sofia picks a positive factor b of B and
3)they write down the product ab on sheet of paper
They repeat the above procedure for all possible order pairs (a, b).At the end, the calculate the of all numbers on the paper is 2340.if both A and B are divisible by 6 and have only 2 and 3 as their prime factors find the least possible value of A+B.

[jaiopetja]

on 5:29
i left 1/(sqrt(x-y)+sqrt(x+y)) then muliti whit (sqrt(x-y)+sqrt(x+y))
...
and on the finish you get dy/dx=-1/2x :D

and when you get x=x(y) you will probably get the same :D

[dr.rahulgupta]

Sir we have third way also : Transfer 1 on LHS and use partial derivatives w.r.t. x a nd y respectively for LHS to get dy/dx .

[avdrago]

I usually like the first method but this time the second method gave us a function in terms of x, so it is better for me

[nikitakipriyanov]

When we got y'=1/2y using the first method, it was very tempting to solve this very easy differential equation. Indeed, the solution is: y²=x+C, but to find out what C is you essentially have to redo everything from the second, explicit method :(

[gergodenes]

I only like implicit differentiation for equations of higher degrees than 2, other than that, regular algebra seems much cleaner.
But I like your explanation for both, so #YAY for me.
Now I was wondering if you could make a video on complex based logarithms, as of a+b*i based logarithm of c+d*i in an e+f*i form. I know the Euler-formula would be great help in this.