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

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

X=0
Then LNY = 0
Y=1
Derivation of X on both sides
[1/(x²+y)]*(x²+y)'=3x²+cosx
(2x+y')/(x²+y)=3x²+cosx
y'=(x²+y)(3x²+cosx)-2x
That is dy / DX = (x? + y) (3x? + cosx) - 2x
x=0,y=1
So dy / DX (x = 0) = 1

Take the derivative y = ln (1 + Ex) and find Dy, which is the x power of E

dy=e^x/(1+e^x) dx

Y = (1 + SiNx) / cosx, is the derivation equal to (SiNx + 1) / (COS ^ 2) x

y=(1+sinx)/cosx
y'=[(1+sinx)/cosx]'
=[cosx*cosx-(1+sinx)*(-sinx)]/cos^2x
=（sinx+1）/cos^2x
You are right! Congratulations!

Derivation y = (2 + cosx) ^ SiNx = e ^ (SiNx * ln (2 + cosx)) how did this part come from? Why did it come out with an e?

For any power exponential function u ^ v = f, take the logarithm and change it into vlnu = LNF
F = e ^ (vlnu)

Find the derivatives of the following functions: （1）y=x2sinx； （2）y=ln（x+ 1+x2）； （3）y=ex+1 ex-1； （4）y=x+cosx x+sinx．

（1）y′=（x2）′sinx+x2（sinx）′=2xsinx+x2cosx．（2）y′=1x+1+x2•（x+1+x2）′=1x+1+x2（1+x1+x2）=11+x2．（3）y′=(ex+1)′(ex-1)-(ex+1)(ex-1)′(ex-1)2=-2ex(ex-1)2．（4）y′=(x+cosx)′(x+sinx)-(x+cos...

Derivation: cosx (1-cos (SiNx))

[ cosx(1-cos(sinx))]'
=-sinx*(1-cos(sinx))+cosx*(cosx*sin(sinx))

What is the derivation of y = ln (x + a)?

y'=1/（x+a）

Derivative y = x ^ LN What's more, how do you get the derivative of X at both ends of the equation when deriving the implicit function I'm so careless... It should be y = x ^ LNX

It seems that you haven't written all the questions, but I can answer your following sentence. When deriving an implicit function, both ends of the equation must remember to take the formula containing y as a compound function about X. for example, we know that the equation E ^ y + Y-1 = e ^ x is an implicit function of y about X, and derivative is required

Derivation, 1) y = 1 / a arccos2x, (2) y = x ^ - 3 + 3 ^ - x (3) y = ln (1-x ^ 2)

1、
y'=1/a*[-1/√(1-(2x)²)]*(2x)'
=-2/[a√(1-4x²)]
2、
y==-3x^(-3-1)+3^(-x)*(-x)'
=-3x^(-4)-3^(-x)
3、
y'=[1/(1-x²)]*(1-x²)
=-2x/(1-x²)

Y = ln [ln (LN x)] derivation

Composite function
f(x)=lnx
g(x)=ln[ln(x)]
r(x)=ln{lnln(x)]}
r'(x)=[1/lnln(x)]g'(x)=[1/lnln(x)][1/ln(x)]f'(x)=[1/lnln(x)][1/ln(x)](1/x)