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# If Sin(x+y)=Log(x+y)then find dy/dx 9 years ago

Sin(x+y)=log(x+y)

Þ   d/dx{sin(x+y)}=d/dx{log(x+y)}

Þ   Cos(x+y)(1+dy/dx)=1/x(1+dy/dx)

Þ   Cos(x+y)(1+dy/dx) - 1/x(1+dy/dx)=0

Þ   (1+dy/dx){ Cos(x+y) – 1/x}=0

Þ   It implies (1+dy/dx)=0

Þ   dy/dx=-1

3 years ago

Sin(x+y)=log(x+y) .  Then dy/dx =??
Step:-1.
d/dx[sin(x+y)]=d/dx[log(x+y)]
Step:-2.
cos(x+y)[1+dy/dx]=1/(x+y)[1+dy/dx]
Step:-3.
cos(x+y)+cos(x+y).dy/dx=1/(x+y)+1/(x+y).dy/dx
Step:-4.
cos(x+y)+cos(x+y).dy/dx-1/(x+y)-1/(x+y).dy/dx=0
Step:-5.
cos(x+y)[1+dy/dx]-1/(x+y)[1+dy/dx]=0
Step:-6.
cos(x+y)-1/(x+y)[1+dy/dx]=0

It is evident from above statement that cos(x+y)-1/(x+y) can be equal to zero but [1+dy/dx] can't be equal to zero.
Hence,

1+dy/dx=0
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