Is the derivative of Xe ^ x-sinx ex cosx

Is the derivative of Xe ^ x-sinx ex cosx

y = xe^x -sinx
y' = (1+x)e^x -cosx

Let z = u ^ 2 + V ^ 2, u = x + y, v = X-Y, find DZ

z'x=z'u*u'x+z'v*v'x
=2u+2v
z'y=z'u*u'y+z'v*v'y
=2u-2v
dz=z'xdx+z'ydy
=(2u+2v)dx+(2u-2v)dy

Let z = f (x / y) and f be differentiable

dz=δf/δxdx+δf/δydy
=f`(x/y)/y*dx+f`(x/y)*(-x/y²)dy
=f`(x,y)(dx/y-xdy/y²)

Let z = f (x, y) find DZ from Z + X + Y - (E Λ x + y + Z) + 2 = 0

Z + X + y-e ^ (x + y + Z) + 2 = 0, x + y + Z = 0 is obviously not satisfied, that is, x + y + Z = 0 ≠ 0
Let's find the partial derivative of X on both sides. Note that z is a function of X and y, and we get the following result
Z '+ 1 - (1 + Z') e ^ (x + y + Z) = 0, the solution is Z '= - 1,
Similarly, Z '= - 1. DZ = - DX dy

Let z = f (x, y) be determined by the equation Z + X + y = e ^ (Z + X + y)

On both sides of the equation, the derivative of X is z'x + 1 = e ^ (Z + X + y) * (z'x + 1), and the solution is z'x = - 1
Two sides of the equation are derived from Y: z'y + 1 = e ^ (Z + X + y) * (z'y + 1), and the solution is: z'y = - 1
So DZ = z'xdx + z'ydy = - DX dy

Let the derivative of function f (x) = x ^ m + ax be f '(x) = 2x + 1, and the sum of the first n terms of sequence {1 / F (n)} (n ∈ n *) be Sn, then the limit of Sn is ()
A、1
B、1/2
C、0
D. It doesn't exist

Let the derivative of function f (x) = x ^ m + ax be f '(x) = 2x + 1, and the sum of the first n terms of sequence {1 / F (n)} (n ∈ n *) be Sn, then the limit of Sn is () a, 1; B, 1 / 2; C, 0; D, and does not exist
F (x) = ∫ (2x + 1) DX = x & # 178; + X + C = x ^ m + ax, so m = 2, a = 1, C = 0, that is, f (x) = x & # 178; + X
1/f(n)=1/(n²+n)=1/[n(n+1)]=(1/n)-1/(n+1)
So S &; n &; = (1-1 / 2) + (1 / 2-1 / 3) + (1 / 3-1 / 4) +. + [1 / n-1 / (n + 1)] = 1-1 / (n + 1)
Therefore, we should choose a

Let f '(x) = 2x + 1, then the sum of the preceding terms of the sequence {1 / F (n)} (n ∈ n *) is

f(x)=x²+x,1/f(n)=1/(n*(n+1))=1/n-1/(n+1)
1/f1+1/f2+.1/f(n)=1-1/(n+1)=n/(n+1).

Finding the derivative of y = (3-x2) / (3 + x2)

(3 + X \\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\#179; - 6x + 2x & #179;) / (3 + X & #178;) & #178; = (- 12x)

The derivative of y = 9-x2
Note: square of X

Solution: y '= 0-2x = - 2x
When x > 0, y 'decreases with the increase of X, so y' decreases monotonically on (0, positive infinity)
When x

The derivative of what is 1 / (1 + x2)

f(x)=arctanx+C
Let y = arctanx; then x = tany
because
f'(x)=(arctanx)'+0
=1/(tany)'
=1/(siny/cosy)'
=1/[(cos^2y+sin^2y)/cos^2y]
=1/(1+tan^2y)
=1/(1+x^2)
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