Y = derivative of sin (3x + 5) How

Y = derivative of sin (3x + 5) How

y'=sin‘(3x+5) × (3x)'=3cos(3x+5)

Finding the derivative of y = sin (3x + 1) is an important process,

f'(x)=lim( Δ x-->0) Δ y/ Δ x=lim( Δ x-->0)[sin(3x 3 Δ x 1)-sin(3x 1)]/ Δ x=lim( Δ x-->0)[2cos(3x 3/2* Δ x 1)sin(3 Δ x/2)]/ Δ x=cos(3x 1)*lim( Δ x-->0)3sin(3 Δ x/2)/(3 Δ X / 2) according to the important limit Lin (X -- > 0) SiNx / x = 1 ≠ Lim（ Δ x-...

When x → 0, Lim [ln (1-2x) + XF (x)] / x ^ 2 = 4, find Lim [f (X-2)] / X If I divide the numerator and denominator by X, I will get Lim [ln (1-2x) / x + F (x)] / x, and then use the Equivalent Infinitesimal Substitution to get the conclusion Lim [f (X-2)] / x = 4. Why is it wrong (the answer is 6)

Answer: 6 solution: Lim_ {x→0}{x[f(x)-2]+2x+ln(1-2x)}/x^2=lim_ {x→0}{x[f(x)-2]}/x^2+lim_ {x → 0} {2x + ln (1-2x)} / x ^ 2 = 4, also Lim_ {x→0}{2x+ln(1-2x)}/x^2=lim_ {x→0}{2x+ln(1-2x)}'/[x^2]'=lim_ {x→0}{2-2/(1-2x)}/2x=lim_ {...

When x → 0, Lim [ln (1 + 2x) + XF (x)] / x ^ 2 = 2, Lim [2 + F (x)] / X requires detailed explanation If I divide the numerator and denominator by X, I will get Lim [ln (1 + 2x) / x + F (x)] / x, and then use the Equivalent Infinitesimal Substitution to get the conclusion 2. Why is it wrong (the answer is 4)

Answer 4 is wrong
Solution 1:
ln(1+2x)~2x
(x→0)   lim [ln(1+2x)+xf(x)]/(x^2)=2
(x→0)   lim [2x+xf(x)]/(x^2)=2
(x→0)   lim [2+f(x)]/x=2
Solution 2: (the defect of this solution is that the title does not say that f (x) is differentiable)

Lim x tends to 0 ln (1 + 2x) / sin3x

What's the equivalent of Robita? It comes out below infinitesimal
Ln (1 + 2x) is equivalent to 2 * x, and sin (3 * x) is equivalent to 3 * x, so that's it

limx[ln(2x+1)-ln(2x)]=?

limx[ln(2x+1)-ln(2x)]
=limx[ln(2x+1)/2x]
=limln[1+1/2x]^x
=limln[1+1/2x]^(2x.1/2)
=limlne^(1/2)
=1/2