The Nth derivative of the square of y = SiNx

The Nth derivative of the square of y = SiNx

y'=2sinxcosx=sin2x
y''=2sin(2x+π/2)
.
y^(n)=2^(n-1)sin(2x+π/2 (n-1))

How to find the derivative of y = sin2x
The derivative of y = SiNx is
y′=cosx
Why is the derivative of y = sin2x
Y ′ = 2cos2x, instead of substituting x for 2x
y′=cos2x

Of course, you can replace x with 2x, but after the replacement, you have to find the derivative of 2x, that is to say, sin2x needs to find the derivative of two functions, one is sin2x = cos2x, the other is 2x = 2, and the product of the two is the derivative

How to find the derivative of y = (sin2x) ^ 2?

y＝(sin2x)^2
y'
=2 sin2x (sin2x)'
=2 sin2x cos2x(2x)'
=4sin2xcos2x
=2sin4x

What is the derivative of y = (sin2x) ^ 3?
The standard answer is 3sin2xsin4x. Let's see where we need to improve?

y=(sin2x)^3
y'=3(sin2x)^2(cos2x)(2)
=6[(sin2x)^2](cos2x)
=3(sin2x)[2(sin2x)(cos2x)]
=3(sin2x)(sin4x)

The derivative of y = e Λ sin2x

y`=(e^sin2x)(sin2x)`(2x)`
=2(e^sin2x)cos2x

The second derivative of y = sin2x-x
RT

y=2cos2x-1

Y = e ^ (- x) * sin2x for second derivative

The beauty of Mathematics
y ' = - e^(-x) * sin(2x) + e^(-x) * 2cos(2x)
= e^(-x) * [ 2cos(2x) - sin(2x) ]
y'' = - e^(-x) [ 2cos(2x) - sin(2x) ] + e^(-x) [ - 4sin(2x) - 2cos(2x) ]
= - e^(-x) * [ 3sin(2x) + 4cos(2x) ]

Find the derivative of the function f (x) = x ^ 2 + 2x + 3 at x = 2

First, the derivative function f '(x) = 2x + 2 of F (x) is obtained
Let x = 2 be substituted into the above formula to get f '(2) = 2 * 2 + 2 = 6

The derivative of F (x) = x ^ 2 - 2x + 3 at x = - 1 is obtained by limit method

f(-1)=(-1)² -2*(-1) +3= 6
f(-1+Δx) =(-1+Δx)² -2*(-1+Δx) +3= Δx² -4Δx+6
f'(-1) = (Δx→0)lim[f(-1+Δx)-f(-1)]/Δx
= (Δx→0)lim[(Δx² -4Δx+6)-6]/Δx
= (Δx→0)lim(Δx -4) =-4;
So f '(- 1) = - 4;

The limit of the first derivative of F (x) is 2 when x tends to infinity, then f (x + k) - f (x) is equal to 2 when x tends to infinity

Mean value theorem: F (x + k) - f (x) = f '(x + AK) * k