y'=0

2.y=x^n

y'=nx^(n- 1)

3.y=a^x

y'=a^xlna

y=e^x

y'=e^x

4.y=logax

y'=logae/x

y=lnx

y'= 1/x

5.y=sinx

y'=cosx

6.y=cosx

y'=-sinx

7.y=tanx

y'= 1/cos^2x

8.y=cotx

y'=- 1/sin^2x

9.y=arcsinx

y'= 1/√ 1-x^2

10.y=arccosx

y'=- 1/√ 1-x^2

1 1.y=arctanx

y'= 1/ 1+x^2

12.y=arccotx

y'=- 1/ 1+x^2

There are several commonly used formulas in the derivation process: 1. y = f [g (x)]，y ' = f '[g(x)]&； 8226; G'(x) in' f' [g(x)], g(x) is regarded as a whole variable, while in G' (x), x is regarded as a variable. "

2.y=u/v,y'=(u'v-uv')/v^2

3. If the inverse function of y = f (x) is x=g(y), then y'= 1/x'

Proof: 1. Obviously, y=c is a straight line parallel to the X axis, so the tangents everywhere are parallel to X, so the slope is 0. Same definition as derivative: Y = C, ⊿ Y = C-C = 0, lim ⊿ x → 0 ⊿ y.

y=e^x

Y' = e x，y=lnx。

The two results of y'= 1/x can be proved by the derivative of composite function. 3.Y = A x，⊿y = a(x+⊿x)-a x = a x(⊿x- 1

⊿y/⊿x=a^x(a^⊿x- 1)/⊿x

If you do ⊿x→0 directly, the derivative function cannot be derived, and an auxiliary function β = a ⊿ x- 1 must be set to substitute for the calculation. From the auxiliary function set, we can know that ⊿x=loga( 1+β). So (a ⊿)

Obviously, when ⊿x→0, β also tends to 0, while Lim β→ 0 (1+β) 1/β = E, so Lim β→ 01/loga (1+β) 65438.

y'=e^x.4.y=logax

⊿y=loga(x+⊿x)-logax=loga(x+⊿x)/x=loga[( 1+⊿x/x)^x]/x

⊿y/⊿x=loga[( 1+⊿x/x)^(x/⊿x)]/x

Because ⊿x→0, ⊿x/x tends to 0 and x/⊿x tends to infinity, lim ⊿ x→ 0 loga (1+⊿ x/x) = logae.

Lim ⊿ x→ 0 ⊿ y/⊿ x = logae/X It can be known that when a=e, there is y=lnx.

Y' = 1/X, and Y = X N can be performed at this time.

Y'= NX(n- 1) is deduced. Because Y = X N and Y = E LN (X N) = E NLNX, Y' = E NLNX &；; 8226; (nlnx)'=x^n&； 8226; n/x=nx^(n- 1).5.y=sinx

⊿y=sin(x+⊿x)-sinx=2cos(x+⊿x/2)sin(⊿x/2)

⊿y/⊿x=2cos(x+⊿x/2)sin(⊿x/2)/⊿x=cos(x+⊿x/2)sin(⊿x/2)/(⊿x/2)

So lim ⊿ x→ 0 ⊿ y/⊿ x = lim ⊿ x→ 0cos (x+⊿ x/2)&; 8226; lim⊿x→0sin(⊿x/2)/(⊿x/2)=cosx

6. Similarly, y=cosx can be deduced.

y'=-sinx.7.y=tanx=sinx/cosx

y'=[(sinx)'cosx-sinx(cos)']/cos^2x=(cos^2x+sin^2x)/cos^2x= 1/cos^2x

8.y=cotx=cosx/sinx

y'=[(cosx)'sinx-cosx(sinx)']/sin^2x=- 1/sin^2x

9.y=arcsinx

x=siny

X' = comfort

y'= 1/x'= 1/cosy= 1/√ 1-sin^2y= 1/√ 1-x^2

10.y=arccosx

X = comfort

x'=-siny

y'= 1/x'=- 1/siny=- 1/√ 1-cos^2y=- 1/√ 1-x^2

1 1.y=arctanx

x=tany

x'= 1/cos^2y

y'= 1/x'=cos^2y= 1/sec^2y= 1/ 1+tan^2x= 1/ 1+x^2

12.y=arccotx

x=coty

x'=- 1/sin^2y

y'= 1/x'=-sin^2y=- 1/csc^2y=- 1/ 1+cot^2y=- 1/ 1+x^2

In addition, in the derivation of complex compound functions such as hyperbolic functions shx, chx, thx, inverse hyperbolic function ARSHX, ARCX, ARTHUX, etc., by consulting the derivation table and using the initial formula and

4.y=u soil v, y'=u soil v'

5.y=uv，y=u'v+uv