Advice: find the derivative of y = x ^ (x ^ 2) by logarithmic derivation, which is the power of X squared,

Advice: find the derivative of y = x ^ (x ^ 2) by logarithmic derivation, which is the power of X squared,

y=x^(x^2)
The natural logarithm of both sides is taken at the same time
lny=(x^2)lnx
Both sides derive x at the same time
y '/y=(x^2) 'lnx+(x^2)·(lnx) '
y '/y=2xlnx+x
y '=y(2xlnx+x)
By substituting y = x ^ (x ^ 2) into the above formula, we can get:
y '=x^(x^2)·(2xlnx+x)

2X plus 5Y minus 3 is equal to 0. Find the x power of 4 times the Y power of 32

2X + 5y-3 = 0, so 2x + 5Y = 3
The x power of 4 can be seen as the 2x power of 2, and the Y power of 32 can be regarded as the 5Y power of 2, so your formula becomes 2x + 5Y power of 2, that is, the third power of 2 is equal to 8. Thank you for adopting it

The derivation of four high numbers: y = arccos (1-2x) y = lncot (x / 2) y = e's minus third power X sin3x y = cos square x times cos (x squared)

1, y '= - 1 / √ [1 - (1-2x) ^ 2] * (- 2) = 2 / √ (4x-4x ^ 2) = 1 / √ (x-x ^ 2) 2, y' = 1 / cot (x / 2) * [- CSC ^ 2 (x / 2)] * 1 / 2 = sin (x / 2) / cos (x / 2) / cos (x / 2) * [- 1 / sin ^ 2 (x / 2)] * 1 / 2 = - 1 / 2 = - 1 / [2Sin (x / 2) cos (x / 2)] = 1 / sin (x / 2) cos (x / 2)] = 1 / sin (sin (x / 2) cos (x / 2)] = 1 / sin (1 / sin (x / 2) cos (x / 2) (x / 2) (x / 2) (x / 2) cos (x / X / 2) = CSC (x / 2) 3

Take the derivative of LN | cosx |

In an interval, for example, cosx > 0 equals - SiNx / cosx = - TaNx
cosx

Dy / DX + Y / x = SiNx / x, x = dispatch time, y = 1, to find the special solution, I use the constant variation method to set u and take the derivative, and then there is u, I will not do it, because there is u 'and u will not be integrated

If u = XY, u ≠ 0, then: Du / DX = y + X (dy / DX) dy / DX = (1 / x) · (DU / DX) - (Y / x). Therefore, the original equation is: (1 / x) · (DU / DX) - (Y / x) = SiNx / XDU / DX = sinxdu = - cosx + C

Y = (x + SiNx) ^ 4 y = x + 1 / X-1 derivative x ^ 2 + y ^ 2-xy = 1 find the derivative of dy / DX implicit function y=1／3x^3。 X = 2 for slope and tangent equation

1、y'=4[（x+sinx）^3]*(1+cosx)
2. Y = x + 1 / X-1 = 1 + 2 / (x-1), and then take your own derivative
3. Take y as a function of X, take the derivative of X on both sides and move to the direction

Let y = f (SiNx), where f is a differentiable function

Well, Dy = DF (SiNx) = f '(SiNx) * D (SiNx) = f' (SiNx) * cosxdx
It turns out that it should be ok here?

Derivative of high number composite function y = ln cos e ^ x, dy / DX

dy/dx=[d(ln cos e^x) / d(cos e^x)] × [d(cos e^x) / e^x] × [d(e^x) / x]
=[1/(cos e^x)] × [- sin e^x] × [e^x]
= - (tan e^x) × e^x

Derivation y = 0.5 (0.5 * ln ((x + 1) / (x-1)) + arctanx)

y=0.5(0.5*ln((x+1)/(x-1))+arctanx)
=0.5(0.5*ln(x+1)-0.5ln(x-1)+arctanx)
y'=0.25/(x+1)-0.25/(x-1)+0.5/(1+x^2)

Let y = f (x) be determined by the equation ln (x ^ 2 + y) = x ^ 3 y + SiNx, and find dy / DX (x = 0)

On both sides, the derivative of X is (2x + dy / DX) / (x ˆ 2 + y) = 3x ˆ 2Y + X ˆ 3dy / DX + cosx
Dy / DX = (3x y + 3x ˆ 2Y ˆ 2 + X ˆ 2cosx + ycosx-2x) / (1-x ˆ 5-x ˆ 3Y)
Substituting x = 0 into ln (x ^ 2 + y) = x ^ 3, y + SiNx, LNY = 0, y = 1
Dy / DX (x = 0) = 1 / 1 = 1, to solve such a problem, we should remember that the result is a constant