Derivative of Square Root Visual: the challenge at timestamp 12:23

[Metronome]

https://www.youtube.com/watch?v=S0_qX4VJhMQ

There is a challenge at 12:23 asking the viewer to arrive at the formula for (d/dx) sqrt(x) by considering small changes to a square. What is the correct way to approach this? I tried to copy the technique used to find (d/dx) (x^2) at 2:25, by letting u = sqrt(x). This makes the area added to the square 2udu + du^2, which comes out to 2sqrt(x)(dsqrt(x)) + (dsqrt(x))^2. The second term is negligible, but how this equals the actual derivative, (1/2)x^(−1/2), is still unclear.

 

[sinx]

https://www.youtube.com/watch?v=S0_qX4VJhMQ

There is a challenge at 12:23 asking the viewer to arrive at the formula for (d/dx) sqrt(x) by considering small changes to a square. What is the correct way to approach this? I tried to copy the technique used to find (d/dx) (x^2) at 2:25, by letting u = sqrt(x). This makes the area added to the square 2udu + du^2, which comes out to 2sqrt(x)(dsqrt(x)) + (dsqrt(x))^2. The second term is negligible, but how this equals the actual derivative, (1/2)x^(−1/2), is still unclear.

I would change the eqn from y=sqrtx to x=y^2; then work with the square of dim y.
see if that helps

 

[Dr.Peterson]

https://www.youtube.com/watch?v=S0_qX4VJhMQ

There is a challenge at 12:23 asking the viewer to arrive at the formula for (d/dx) sqrt(x) by considering small changes to a square. What is the correct way to approach this? I tried to copy the technique used to find (d/dx) (x^2) at 2:25, by letting u = sqrt(x). This makes the area added to the square 2udu + du^2, which comes out to 2sqrt(x)(dsqrt(x)) + (dsqrt(x))^2. The second term is negligible, but how this equals the actual derivative, (1/2)x^(−1/2), is still unclear.

The derivative is ∆y/∆x, where in this case y = √x. You've seen that ∆x = 2y∆y + ∆y^2. Neglecting that second term, ∆x ≈ 2y∆y. Rearranging, ∆y ≈ ∆x/(2y), so that ∆y/∆x ≈ 1/(2y) = 1/2 x^(-1/2).