Let f (x) = e ^ (3x), find f '(x) by using the definition of derivative This is mainly about derivatives

Let f (x) = e ^ (3x), find f '(x) by using the definition of derivative This is mainly about derivatives

f'(x)=lim(h→0)(e^(3x+3h)-e^(3x))/h
=e^(3x)*lim(h→0)(e^(3h)-1)/(3h)*3
=e^(3x)*1*3
=3e^(3x)

What is the derivative of [1 / (1 + x)]

Ln (1 + x) derivative equals [1 / (1 + x)]

Whose derivative is 1 / x

Namely seeking
∫1/xdx
=lnx +C

The derivation of a compound function Find the derivative of F (x) = ln (x + 1) / X

f(x)=ln[(x+1)/x]
f'(x)=x/(x+1)*[(x+1)/x]'
=x/(x+1)*(-1/x ²)
=-1/(x ²+ x)
f(x)=[ln(x+1)]/x
f'(x)=[x/(x+1)-ln(x+1)]/x ²
=1/(x ²+ 1)-[ln(x+1)]/x ²

Derivative problem of composite function f(x)=2÷(x ²+ 11) How to find the derivative? Do you use the derivative method of composite function? How to use it?

This requires the use of compound functions and the four Operational derivation rules of functions
Note u = x ²+ 11, then u '= 2x
f(x)=2/u
f'(x)=-2/u ²* u'=-2/u ²* 2x=-4x/(x ²+ 11) ²

Find the derivative, (10) y=ln√x+√lnx (12)y=sinnx (14)y=sin^nx (16)y=cos^3 x/2 (18) y=lntanx/2 (20) y=lnlnx (22)y=1/(cos^n x) (24)y=sec^2 x/a+csc^2 x/a

(10) y=ln√x+√lnx=ln(x^0.5)+(lnx)^0.5=0.5*ln x+(lnx)^0.5
y'=0.5/x+0.5*(lnx)^(-0.5)/x=(0.5/x)*(1+1/((lnx)^0.5)
(12)y=sinnx
y'=(cosnx)*n=n*(cosnx)
(14)y=sin^nx=(sinx)^n
y'=n*(sinx)^(n-1)*cosx=n*cosx*(sinx)^(n-1)
(16)y=cos^3 x/2=(cos x/2)^3
y'=3*(cos x/2)^2*(sin x/2)/2=(3/2)*(sin x/2)*(cos x/2)^2
(18) y=lntanx/2=ln(tan(x/2))
y'=1/(tanx/2)*(sec(x/2))^2/2=(sec(x/2))^2/(2*tanx/2)=sin(x/2)/(2*(cos(x/2))^3)
(20) y=lnlnx=ln(lnx)
y'=1/(lnx)/x=1/(x*lnx)
(22)y=1/(cos^n x)=1/(cos x)^n=(cos x)^(-n)
y'=-n*(cos x)^(-n-1)*(-sin x)=(n*sin x)/((cos x)^(n+1))
(24)y=sec^2 x/a+csc^2 x/a=(sec x/a)^2+(csc x/a)^2
y'=2*(sec x/a)*(tan x/a)*(sec x/a)/a=(2/a)*(tan x/a)*(sec x/a)^2=(2*sin x/a)/((cos x/a)^3)