find dy/dx in sin square y + cos xy = 27
Answers
Answer:
Consider the given equation.
sin
2
y+cosxy=π
Differentiate both sides with respect to x.
2siny×cosy×
dx
dy
−sinxy×(y+x
dx
dy
)=0
2sinycosy
dx
dy
−ysin(xy)−xsin(xy)
dx
dy
=0
2sinycosy
dx
dy
−xsin(xy)
dx
dy
=ysin(xy)
(sin2y−xsin(xy))
dx
dy
=ysin(xy)
dx
dy
=
sin2y−xsin(xy)
ysin(xy)
Hence, this is the required result.
Explanation:
I hope it helps you dear ❣️❣️☺️☺️❤️❤️
Answer:
Consider the given equation.
sin
2
y+cosxy=π
Differentiate both sides with respect to x.
2siny×cosy×
dx
dy
−sinxy×(y+x
dx
dy
)=0
2sinycosy
dx
dy
−ysin(xy)−xsin(xy)
dx
dy
=0
2sinycosy
dx
dy
−xsin(xy)
dx
dy
=ysin(xy)
(sin2y−xsin(xy))
dx
dy
=ysin(xy)
dx
dy
=
sin2y−xsin(xy)
ysin(xy)
Hence, this is the required result.
Explanation:
I hope it helps you dear ❣️❣️☺️☺️❤️❤️
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