Y = (x / LNX) + sin e find dy The answer is (lnx-1) / ln ^ 2x. I don't understand what the derivative of sin e is, not cos

Y = (x / LNX) + sin e find dy The answer is (lnx-1) / ln ^ 2x. I don't understand what the derivative of sin e is, not cos

E in sine is a constant of 2.71828, so sine is also a constant, and the derivative is 0. What you call SiNx, where x is the independent variable of a function

Let y = LNX + SiN x / 1. Find dy

y'=1/x-cosx/sin^2x
dy=(1/x-cosx/sin^2x)dx

Y = x * sin (LNX) find dy

y=x*sin(lnx)
y'=sin(lnx) +x*cos(lnx)*(lnx)'
=sin(lnx) +x*cos(lnx)*1/x
=sin(lnx) +cos(lnx)
dy=[sin(lnx) +cos(lnx)]dx

Find the differential equation. # replace the root sign xydx + # (1-x ^ 2) dy = 0

xydx+√(1-x^2) dy=0√(1-x^2)dy=-xydx-dy/y=xdx/√(1-x^2)-dy/y=dx^2/2√(1-x^2)2dy/y=d(1-x^2)/√(1-x^2)2lny=(-1/2+1)(1-x^2)^(-1/2+1)=1/2 *√(1-x^2)+C4lny=√(1-x^2)+Clny=√(1-x^2)/4+C/4y=e^(√(1-x^2)+C/4)...

Find the general solution of differential equation (1 + x square) dy + xydx = 0

(1+x ²) dy+xydx=0
==>dy/y=-xdx/(1+x ²)
==>ln│y│=(-1/2)ln(1+x ²)+ Ln │ C │ (C is the integral constant)
==>y=C/√(1+x ²).

The general solution of the differential equation xydx + (1 + x ^ 2) dy = 0 is y= A、y^2=C/1+x^2 B、y=C/1+x^2 C、y^2=C/1+x D、y=C/1+x

xydx+(1+x^2)dy
→(1/2)·[1/(1+x^2)]dx^2+(1/y)dy＝0
∴(1/2)ln(1+x^2)+lny+C＝0.
It can also be expressed as: y ^ 2 = C / (1 + x ^ 2)