How do you find the derivative of #x+tan(xy)=0#?

2 Answers
Jan 21, 2017

# dy/dx={(1+x^2)arc tanx-x}/{x^2(1+x^2)}#

Explanation:

#x+tan(xy)=0 rArr tan(xy)=-x rArr xy=arc tan(-x)#

# rArr xy=-arc tanx#

# y=-(arc tanx)/x#

By the Quotient Rule , then,

#dy/dx=-[{xd/dx(arc tanx)-(arc tanx)d/dx(x)}/x^2]#

#=-[{x/(1+x^2)-arc tanx}/x^2]#

#:. dy/dx={(1+x^2)arc tanx-x}/{x^2(1+x^2)}#

Jan 21, 2017

#dy/dx=- (cos^2 (xy)+y)/x#

Explanation:

Use implicit differentiation and algebra to get #dy/dx# on one side of the equation

  1. Differente both sides of the equation
    #color(red)(d/dx)(x)+color(red)(d/dx)(tan(xy))=color(red)(d/dx)(0)#

  2. #d/dx(x)=1# and #d/dx(0)=0#
    #color(red)(1)+d/dx(tan(xy))=color(red)(0)#

  3. The derivative of the #tan# function is #sec^2#
    #d/dx(tan (xy))=sec^2 (xy)*(d (xy))/dx# (Chain Rule)
    #1+color(red)(sec^2(xy)(d(xy))/dx)=0#

  4. Apply the Product Rule:
    #(d(xy))/dx=x dy/dx+y dx/dx=x dy/dx+ycancel(dx)/cancel(dx)#
    #1+sec^2 (xy)*color(red)((x dy/dx + y))=0#

  5. Subtract 1 from both sides
    #sec^2 (xy)*(x dy/dx + y)=color(red)(-1)#

  6. Divide both sides by #sec^2 (xy)#
    #x dy/dx + y = -1/color(red)(sec^2 (xy))#

  7. #1/sec^2 (xy)=cos^2(xy)#
    #x dy/dx + y = -color(red)(cos^2 (xy))#

  8. Subtract #y# from both sides
    #x dy/dx=-cos^2(xy)color(red)(-y)#

  9. Divide both sides by #x#
    #dy/dx=(-cos^2(xy)-y)/color(red)(x)#

  10. Simplify
    #dy/dx=color(red)(-) (cos^2(xy)color(red)(+)y)/x#