N-th derivative of y = e ^ x

N-th derivative of y = e ^ x

Or y = e ^ x

Find the n-order derivative of e ^ (- 1 / x ^ 2). Another definition: y = 0 when x = 0

This problem is a little different, because (0,0) can not determine whether n-order differentiable, so we must use the definition of derivative to solve it
(first, the constant (x → 0) Lim e ^ (- 1 / x ^ 2) / x ^ k = 0
It is proved that when k ≤ 0, it is obviously true
k> When 0, (x → 0) Lim e ^ (- 1 / x ^ 2) / x ^ k = (t →∞) Lim T ^ k / e ^ (T ^ 2) = 0, the result can be obtained by using lobida's law continuously)
The n-th derivative is represented by Y (n)
Derivative at x = 0
y(1)=lim [e^(-1/x^2)-0]/(x-0)=0
y(2)=lim 2/x^3*e^(-1/x^2)/x=0
No matter how many derivatives are obtained, the n-order derivative, when x ≠ 0, always contains e ^ (- 1 / x ^ 2), so
y(n)=0
Then the general situation
e^（-1/x^2)=y
-1/x^=lny
Derivation
2/x^3=y'/y
2y=x^3y'
Finding n-order derivative by Leibniz formula
2y(n)=y(n+1)+3nx^2y(n)+3n(n-1)xy(n-1)+n(n-1)(n-2)y(n-2)
So get
y(n+1)+y(n)(3nx^2-2)+3n(n-1)xy(n-1)+n(n-1)(n-2)y(n-2)=0
Fourth order homogeneous linear differential equation
There is no general solution for second order homogeneous linear differential equations
Halo is generally to find the n-th derivative of a special point

Solving n-order derivative: y = e ^ ((- x) ^ (- 2)) urgent

Halo...... do this problem for the fourth time. Is it to find the n-order derivative at any point or the n-order derivative at the origin? If the latter is included, there must be y = 0 at the origin, otherwise the n-order derivative does not exist. (first, it is proved that there is always (x → 0) Lim e ^ (- 1 / x ^ 2) / x ^ k = 0: when k ≤ 0, it is obvious that when k > 0, (x → 0) Lim e ^ (- 1 / x ^ 2) / x ^ k = (t →∞

Find the - x derivative of E The process must be detailed y'=e^[e^(-x)]*u'*v' How did you get here

Let y = e ^ [e ^ (- x)]
Let v = - x, u = e ^ V, t = e ^ U
y'=t'*u'*v'
= e^[e^(-x)]*u'*v'
=e^[e^(-x)]*[e^(-x)]*(-1)
=-[e^(-x)]*e^[e^(-x)]
Don't know how to ask me!

Find the derivative! Find the derivative of e ^ xcose ^ x,

This is a composite function, t = e ^ x, and the original function f (T) = tcost, then:
Derivative of F = f ` (T) × t`(x)=(cost-tsint)(e^x)=(cose^x-e^xsine^x)(e^x)

Find the derivative of derivative (x Λ (E Λ x))

y = x^(e^x)
lny = (e^x)lnx
y'/y = (e^x)lnx + (e^x)/x = (e^x)(lnx + 1/x)
y' = y(e^x)(lnx + 1/x)
= (e^x)*x^(e^x)(lnx + 1/x)