How to get the derivative of x power + 1 of y = 4 / E

How to get the derivative of x power + 1 of y = 4 / E

y=4(e^x+1)^(-1)
So y '= - 4 (e ^ x + 1) ^ (- 2) * (e ^ x + 1)'
=-4e^x/(e^x+1)²

Derivation of e to the negative x power?

=-e^(-x)

The Y power of E | + xy-e = 0 belongs to the simplest implicit function derivation!

y'e^y+y+xy'=0
y'=-y/(x+e^y)

Why is the result of E's y-th power + xy-e's derivative for X is: (E's y power times dy / DX) + y + x times dy / DX And my result is: (E's y power) + y + x times dy / DX, what's wrong with me? Please explain it in detail, I'm poor in math

z=e^y+xy-e
z' |x =y'e^y+(y+xy')
One problem with your result is that e ^ y is a composite function, and the derivative of y to X is followed by the derivative: y '

Calculation: (x - 1 power - Y - 1 power) / (x - 2 - Y - 2) - XY (x + y) - 1

The original formula = (1 / X-1 / y) / (1 / x? - 1 / y? - XY (x + y)
=(1/x-1/y)÷[(1/x-1/y)(1/x+1/y)]-xy(x+y)
=1/(1/x+1/y)-xy(x+y)
=1/[(x+y)/xy]-xy(x+y)
=xy/(x+y)-xy(x+y)
=xy*[1-(x+y)²]/(x+y)
=xy(1-x-y)(1+x+y)/(x+y)

What's the 3x power derivative of E? RT

（e^3x）’=（e^3x）*（3x）’=3e^3x.

To the sixth power of (3x-5), how much is a derivative, Can we start with logarithms,

No, according to the first derivative formula of (AX + b) ^ m = ma (AX + b) ^ M-1
The reciprocal of (3x-5) ^ 6 is 18 (3x-5) ^ 5

Let f (x) = SiNx + cosx, f '(x) be the derivative of F (x), if f (x) = 2F' (x), then sin2x − sin2x cos2x=______ .

∵f（x）=sinx+cosx，
∴f′（x）=cosx-sinx
∵f（x）=2f′（x）
ν SiNx + cosx = 2 (cosx SiNx), i.e. 3sinx = cosx
∴sin2x−sin2x
cos2x=sin2x−2sinxcosx
cos2x=sin2x−6sin2x
9sin2x=−5
Nine
So the answer is: − 5
Nine

If the derivative of F (x) is cosx, then the original function of F (x) is () a: 1 + SiNx B: 1-sinx C: 1 + cosx D: 1-cosx

A
The basic definition of derivative
However, the title is not rigorous. It should be "the original function may be"

If the derivative of F (x) is SiNx, then f (x) has an original function () A. 1+sinx B. 1-sinx C. 1+cosx D. 1-cosx

The derivation of the four options can be found to be SiNx
For a, (1 + SiNx) ′ = cosx, exclude a;
For B, (1-sinx) ′ = - cosx, B is excluded;
For C, (1 + cosx) ′ = - SiNx, C is excluded;
From the exclusion method, D can also be verified, (1-cosx) ′ = SiNx, which meets the requirements
Therefore, D