How to find y = e ^ sin2x dy

dy=(e^(1/x)(-1/x^2)+cos2x*2)dx y=e^(1/x)+sin2x dy=e^(1/x)d(1/x)+cos2xd2x =e^(1/x)(-1/x^2)dx+2cos2xdx =[-e^(1

Let y = 1 / x power of E + sin2x, find dy

y=e^(1/x)+sin2x
dy=e^(1/x)d(1/x)+cos2xd2x
=e^(1/x)(-1/x^2)dx+2cos2xdx
=[-e^(1/x)/x^2+2cos2x]dx

RELATED INFORMATIONS

1. Dy / DX + (2 / x) y = - x when y (2) = 0,
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15. What is ∫ = dy / DX? How to apply it? What are the related formulas?
16. The general solution formula of the first order linear differential equation (X-2) * dy / DX = y + 2 * (X-2) ^ 3, find the general solution of Y ∵(x-2)*dy/dx=y+2*(x-2)³ ==>(x-2)dy=[y+2*(x-2)³]dx ==>(x-2)dy-ydx=2*(x-2)³dx ==>[(x-2)dy-ydx]/(x-2)²=2*(x-2)dx ==>d[y/(x-2)]=d[(x-2)²] ==>y/(x-2)=(x-2)² +C & nbsp; (C is an integral constant) & nbsp; & nbsp; & nbsp; = = & gt; y = (X-2) & # 179; + C (X-2) & nbsp; & nbsp; & nbsp;; the general solution of the original equation is y = (X-2) & # 179; + C (X-2) & nbsp; (C is an integral constant) How to get the following from the above formula?
17. The solution dy / DX + 1 / 3Y = 1 / 3 * (1-2x) y ^ 4 I feel the answer is wrong,
18. Try to derive from DX / dy = 1 / y ': D ^ 2x / dy ^ 2 = - y' '/ (y') ^ 3 the problem of D [1 / y '] / DX} * [DX / dy] D [1 / y '] / DX} * [DX / dy] (because y' in 1 / y 'is the derivative of the function y = f (x), it is the function of X, so 1 / y' is also the function of X. if the function of X wants to derive y now, it needs the derivative method of composite function. For 1 / y ', it needs to derive x first, and then y) Since y is a function of X, why does 1 / y 'need to derive x from y after deriving x? Is it because the inverse function x is also a function of Y A little dizzy
19. The proof of dy / DX f（x）＝cos x＋sin x／cos x－sin x It is proved that f '(x) = 2 / 1-sin 2x . .
20. For any real number x, y ∈ R, SiNx + cosy = 2Sin (X-Y / 2 π / 4) cos (x y / 2 - π / 4), then sin13 π / 24. Cos5 π / 24 = =? Which formula is used in this problem? I don't want it. I need the formula