Let y = f (x) be determined by the equation ln (x + y) = XY ^ 2 + SiNx, then dy / dx|x = 0 =? How to calculate it

Let y = f (x) be determined by the equation ln (x + y) = XY ^ 2 + SiNx, then dy / dx|x = 0 =? How to calculate it

Put x = 0 into the equation and get y = 1,
Then use the implicit function derivation rule, both sides of the X derivative (y can be replaced by F (x), so as to avoid mistakes)
There is,
On the left is (1 + y ') / (x + y)
On the right is y ^ 2 + 2xyy '+ cosx
Replace x = 0, y = 1 into
thus
(1+y')/1=1+1
Y '= 1,
that is
dy/dx|x＝0＝1

Let y = y (x) be determined by the equation ln (x2 + y) = x3y + SiNx, then dy dx|x＝0=______ .

The derivation of X on both sides of the equation is obtained
2x+y′
x2+y＝3x2y+x3y′+cosx
y′＝2x−(x2+y)(3x2y+cosx)
x5+x3y−1
According to the original equation, when x = 0, y = 1, substituting into the above formula, we can get
y′|x＝0＝dy
dx|x＝0＝1
So the answer is: 1

Let y = y (x) be determined by the equation ln (x ^ 2 + y) = x ^ 3Y + SiNx, then dy | (x = 0)

Substitute x = 0 to get y = 1
The derivative of X on both sides
(2x+y')/(x^2+y)=3x^2y+x^3y'+cosx
X = 0, y = 1
y'=1
dy|（x=0） =dx

Let y = y (x) be determined by the equation ln (x ^ 2 + y ^ 2) ^ 1 / 2 = arctany / x, and find dy / DX

See figure

X (to the power of Y) = y (to the power of x) to find dy / DX by deriving two sides with implicit function It is required that the derivative method of implicit function without partial derivative should be used, and the ordinary method of deriving two sides should be used,

ylnx=xlny
y/x+y'lnx=lny+(x/y)y'
y'=(lny-y/x)/(lnx-(x/y))

Y = ln (x power of 1 + e) to find dy

y'=[1/(1+e^x)]*e^x

How to find dy of y = ln (4th power) (1-x)

The derivative of y = [ln (1-x)] ^ 4Y = 4 [ln (1-x)] ^ 3 * 1 / (1-x) * (- 1) = 4 / (1-x) * [ln (1-x)] ^ 3dy = 4 / (1-x) * [ln (1-x)] ^ 3DX compound function, and then take the derivative from outside to inside

Y = e ^ x (cosx + SiNx) derivation

E ^ X and derivatives in brackets
y'=e^x(cosx+sinx)+e^x*(-sinx+cosx)=2cosx*e^x
In (), it is regarded as the coefficient of e ^ X

Y = e ^ SiNx + (SiNx) ^ cosx derivation

y=e^sinx+e^(cosxlnsinx)
y'=e^(sinx) cosx+e^(cosx lnsinx)(cosx cosx/sinx -sinxlnsinx)

Derivation of y = x SiNx cosx

y'=(xsinx)'-(cosx)'
=x'sinx+ x(sinx)'-(cosx)'
=sinx+xcosx+sinx
=2sinx+xcosx