Y = 2x ^ 3-x ^ 2 + 5, then the fifth derivative of Y is?

Y = 2x ^ 3-x ^ 2 + 5, then the fifth derivative of Y is?

y1=6x^2-2x
y2=12x-2
y3=12
y4=0
y5=0

Finding the n-order derivative of y = (x ^ 3-1) / (x ^ 2-2x-3)
In fact, I also understand how to change it into 1 / (AX + b), but the question is, how can we get to this point?
Just tell me how (x ^ 3-1) / (x ^ 2-2x-3) turns into x + 2 + (7x + 5) / (x + 1) (x + 3), and I'll do it myself

So the original formula = x + (AX ^ 2 + BX + C) / (x ^ 2-2x-3) = (x ^ 3 + (A-2) x ^ 2 + (B-3) x + C) / (x ^ 2-2x-3) so a = 2, B = 3, C = - 1, so the original formula = x + (AX ^ 2 + BX + C) / (x ^ 2-2x-3) = x + (2x ^ 2 + 3x-1) / (x ^ 2-2x-3) similarly, the original formula = x + 2 + (7x + 5) / (x

Finding the n-order derivative of y = (x ^ 2 + 2x + 2) e ^ (- x)

F (x) = x & # 178; + 2x + 2, f '(x) = 2 (x + 1), f' '(x) = 2g (x) = e ^ (- x), the n-order derivative of G (x) is: (- 1) ^ n * e ^ (- x) using Leibniz formula y ^ (n) = f (x) * G ^ (n) (x) + C (n, 1) * f' (x) * G ^ (n-1) (x) + C (n, 2) * f '' (x) * G ^ (n-2) (x) =

The second derivative of y = 1 / x ^ 2-2x-8
The second derivative of y = 1 / (x ^ 2-2x-8), leaving out the brackets. Ha ha

y（1）=-2X^(-3)-2
y（2）=6X^（-4）

Derivative of log3cos ^ 2x

Let u = cosx v = u ^ 2
(log3 v)'=(1/v*ln3)*u'*v'=(1/u^2*ln3)*(-sin x)*2u=(1/cos^2x*ln3)*(-sin x)*2cos x
Let's tidy up the rest

Consult the derivative of log3cos ^ 2x
Please explain the solution,

Using the derivative rule of compound function
Log3f (x) derivative is to find the first derivative of logarithm and then multiply it by the derivative of F (x). If f (x) is also a composite function, then enter the inner layer, this is the chain derivation rule of composite function
Specifically: [log3 (COS ^ 2 x)] '
=1/(cos^2 x)ln3 * (cos^2 x)'
=1/(cos^2 x)ln3 * (-2cosxsinx)
=-2tanx/ln3

How to calculate the derivative of e ^ - x
Why do I use 1 / e ^ x instead of compound derivative
What's wrong with me

Consider - x as a composite function,
Then (e ^ - x) '= e ^ - x (- x)' = - e ^ - X
It's OK to convert to 1 / e ^ x first, and treat e ^ x as a composite function
(1/e^x)'=-1/(e^x)^2*(e^x)'=-1/e^x=-e^-x

Find the second derivative of the function, y = x ^ 2E ^ X

y'=2xe^x+x^2e^x
y''=2e^x+2xe^x+2xe^x+x^2e^x=2e^x+4xe^x+x^2e^x

Finding the derivative of function f (x) = x ^ 2E ^ - x

Take the natural logarithm on both sides, where ln y = (2e ^ - x) * ln x, and take the derivative on both sides of the equation, where y '/ y = 2 ((- e ^ - x) * ln X
+(e ^ - x) / x), so f '(x) = y' = 2 ((- e ^ - x) * ln x + (e ^ - x) / x) * y, that is, the derivative is 2 ((- e ^ - x) * ln x + (e ^ - x) / x) * (x ^ 2E ^ - x)

who taught you ____ a cat?A.feed B.feeding C.to feed D.fed

C
Teach sb to do sth