Asked by marcus
find dy/dx of xsiny+cos2y=cosy
Answers
Answered by
drwls
Use implicit differentiation.
Differentiate both sides of the equation with respect to x, treating y as a function of x.
x cosy dy/dx + sin y - 2 sin 2y dy/dx = -sin y dy/dx
Now solve for dy/dx. It will be a function of both x and y.
You could also differentiate both sides with respect to y, solve for dx/dy, and take the reciprocal of the answer. The result will look different but will still be correct.
x cos y + sin y dx/dy - 2 sin 2y = -sin y
dx/dy = [-siny + 2 sin2y -x cos y]/sin y
dy/dx = sin y/[-siny + 2 sin2y -x cos y]
Differentiate both sides of the equation with respect to x, treating y as a function of x.
x cosy dy/dx + sin y - 2 sin 2y dy/dx = -sin y dy/dx
Now solve for dy/dx. It will be a function of both x and y.
You could also differentiate both sides with respect to y, solve for dx/dy, and take the reciprocal of the answer. The result will look different but will still be correct.
x cos y + sin y dx/dy - 2 sin 2y = -sin y
dx/dy = [-siny + 2 sin2y -x cos y]/sin y
dy/dx = sin y/[-siny + 2 sin2y -x cos y]
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