What is the implicit derivative of #25=cosy/x-3xy#?

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Answer 1 · 2017-03-17T23:05:34

#dy/dx = -(25 +6xy)/(siny+3x^2)#

Differentiate both sides of the equation with respect to #x#:

#d/dx(cosy/x-3xy) = 0#

# (-xsinydy/dx-cosy)/x^2 -3(y+xdy/dx) = 0#

Since #x != 0# we can multiply both sides by #x#:

# -sinydy/dx-cosy/x -3xy-3x^2dy/dx = 0#

#dy/dx(siny+3x^2) = -cosy/x -3xy#

From the original expression we can substitute:

#cosy/x = 25+3xy#

so:

#dy/dx(siny+3x^2) = -25 -6xy#

and finally:

#dy/dx = -(25 +6xy)/(siny+3x^2)#