問數學(derivative)

y^2 ln(x+1) - y/(e^x+sinx) + 2cosx.ln(x^2+1) = 1

Differentiate w.r.t. x
y^2/(x+1) + 2yln(x+1) dy/dx -(dy/dx)/(e^x+sinx)+y*(e^x+cosx)/(e^x+sinx)^2 + 4xcosx/(x^2+1)-2sinxln(x^2+1) = 0
dy/dx =[y^2/(x+1)+y*(e^x+cosx)/(e^x+sinx)^2 + 4xcosx/(x^2+1)-2sinxln(x^2+1)]/[2yln(x+1)-1/(e^x+sinx)]

When x = 0,
y^2 ln(0+1) - y/(e^0+sin0) + 2cos0.ln(0^2+1) = 1
0-y+0 =1
y = -1
Substitute x = 0 and y = -1 into dy/dx
dy/dx at x = 0 is [1/1+ (-1)(1+1)/(1+0)^2 + 4*0*1/1 - 2*0*0] / (2(-1)(0) - 1/(1+0))
dy/dx at x = 0 is (1-2)/(-1) = 1

y^2 ln(x+1) - y/(e^x+sinx) + 2cosx.ln(x^2+1) = 1

when x=0,
y^2 ln(0+1) - y/(e^0+sin0) + 2cos0.ln(0^2+1) = 1
y^2*0 - y + 2*0 = 1
y= -1

differentiate, y^2 ln(x+1) - y/(e^x+sinx) + 2cosx.ln(x^2+1) = 1, w.r.t. x

2y*dy/dx*ln(x+1)+ y^2*[1/(x+1)]
-dy/dx/[(e^x+sinx) ] +y*(e^x+cosx)/(e^x+sinx)^2
- 2sinx.ln(x^2+1) + 2cosx*(2x)[1/(x^2+1)] =0

2y*dy/dx*ln(x+1)+ y^2/(x+1)
-dy/dx/[(e^x+sinx) ] +y*(e^x+cosx)/(e^x+sinx)^2
- 2sinx.ln(x^2+1) + 4xcosx/(x^2+1) =0 ...(1)

put x=0, y=-1 into (1)

2(-1)*dy/dx*ln(0+1)+ (-1)^2/(0+1)
-dy/dx/[(e^0+sin0) ] +(-1)*(e^0+cos0)/(e^0+sin0)^2
- 2sin0.ln(0^2+1) + 4(0)cos0/(0^2+1) =0

-2dy/dx*(0)+ (-1)^2
-dy/dx/[(1+0) ] +(-1)*(1+1)/(1+0)^2
- 2(0).ln(0+1) + 4(0)cos0/(0+1) =0

0+ 1-dy/dx -2-0 + 0 =0
-dy/dx -1=0
dy/dx =-1