Why is the differential symbol used by Leibniz better than Newton's? Please give me an example. Thank you

Why is the differential symbol used by Leibniz better than Newton's? Please give me an example. Thank you

Newton's symbol is a dot on the letter
Leibniz's symbol is dy / DX
In this way, to express the higher derivative, Newton's sign has to add a lot of dots on the letter, while Leibniz's sign only needs to write d ^ N Y / DX ^ n, plus an exponent
A little bit more than you can count

(root sign 6-2 times root sign 15) multiply root sign 3-6 times root sign 1 / 2 to write the process. The mathematical symbol of the process is in Chinese characters

=(√6－2√15)×(√3)－6√(1/2)
=√18－2√45－3√2
=3√2－6√5－3√2
=－6√5

(2) under the root sign, x + 5 = x + 1, x =? Ah
It's better to write down the process as well
Under the root sign above is (x to the second power + 5)
The root sign (x2 power + 5) = x + 1, x =?

X + 5 = x + 1 under (2 times) radical
√X+5=X+1
X+5=(X+1)^2
X^2+2X+1=X+5
X^2+X-4=0,
X=(-1+√17)/2

1-radical 1 / 2 =? (there should be a calculation process. It's better to use numbers and symbols instead of Chinese characters). I'm in a hurry. I hope there will be an answer this morning,
Root 3 minus root 1 / 2 is equal to? (same as above) thank you

1-√1/2=1- √2 /2=1-0.707=0.293
√3-√1/2=√3- √2 /2=1.732-0.707=1.025

Y = x ^ 2-2xsinx ^ 3

dy=d(x^2)-2d(xsinx^3)
=2xdx-2(sinx^3)dx-2xd(sinx^3)
=(2x-2sinx^3)dx-2xcosx^3d(x^3)
=(2x-2sinx^3)dx-2xcosx^3·3x^2dx
=(2x-2sinx^3-6x^3·cosx^3)dx

If f (x, y) is differentiable at (x0, Y0), then its X, y partial derivatives are continuous at (x0, Y0)?

Wait a minute. Write it right away

The solution of inequality (a + b) x + (2a-3b) less than 0 is x greater than 3. What is the solution of inequality (a + b) x + (2a-3b) greater than 0?

(a+b)x+(2a-3b)0
bx>3b
b

The solution set of inequality (2a-3) x + A + 1 > 0 is X

∵(2a-3 )x+a+1>0,
　　∴(2a-3 )x>-a-1
Its solution set is X

Solving inequality: 2x ^ 2 - (2a-1) x-a

(2x+1)(x-a)

It is known that the system of inequalities x + 1 / X-1 ≥ 0 on X, ① (x-2a + 1) (x-a ^ 2) ≤ 0, ② where a ∈ R
(1) if the solution set of the inequality system is an empty set, find the range of A
(2) If the solution set of inequality system is a nonempty set {x | B ≤ x ≤ - 1}, the value range of real number B is obtained
Process details~

If x ＞ 1 or X ≤ - 1 is obtained from (1), then (2) is sorted into (x - (2a-1)) (x-a ^ 2), and the two ends are 2a-1 and a ^ 2, because a ^ 2 - (2a-1) ≥ 0. If a = 1, the solution set is empty. If a ≠ 1, because 2a-1 > - 1 and a ^ 2