1.y=c(c is a constant) 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 = Tanks Y' =1/cos 2x
8.y=cotx y'=- 1/sin^2x
9 . y = arcsinx y'= 1/√ 1-x^2
10 . y = arc cosx y'=- 1/√ 1-x^2
1 1 . y = arctanx y'= 1/ 1+x^2
12 . y = arccotx y'=- 1/ 1+x^2
In the process of derivation, there are several commonly used formulas to be used:
1.y=f[g(x)]，y'=f'[g(x)]？ 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'
Certificate: 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. The definition of derivative is the same: y=c,⊿y=c-c=0,lim⊿x→0⊿y/⊿x=0.
2. The derivation of this is not proved for the time being, because if it is deduced according to the definition of derivative, it cannot be extended to the general case that n is an arbitrary real number. Two results, y = e x y' = e x and y=lnx y'= 1/x, can be proved by the derivative of composite function.
3.y=a^x，
⊿y=a^(x+⊿x)-a^x=a^x(a^⊿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. We can know from the auxiliary function: ⊿x=loga( 1+β).
So (a ⊿ x-1)/⊿ x = β/loga (1+β) =1/loga (1+β)1/β.
Obviously, when ⊿x→0, β also tends to 0. And lim β→ 0 (1+β) 1/β = e, so lim β→ 01/loga (1+β)1/logae =
Substituting this result into lim⊿x→0⊿y/ ⊿ x = lim ⊿ x → 0ax (a ⊿ x-1)/⊿ x gives lim ⊿ x → 0 ⊿ y/.
We can know that when a=e, there is y = e x 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 = lnxy' =1/x.
At this point, y = x n y' = NX (n- 1) can be deduced. Because y = x n, y = e ln (x n) = e nlnx,
So you're NLNX? (nlnx)'=x^n？ 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)? lim⊿x→0sin(⊿x/2)/(⊿x/2)=cosx
6. Similarly, y=cosx y'=-sinx can be deduced.
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 '
You can get results quickly.