What is the implicit derivative of #1=ytanx-y^2#?

1 Answer
Dec 19, 2015

#dy/dx=(ysec^2x)/(2y-tanx)#

Explanation:

#d/dx[1=ytanx-y^2]#

Remember that taking the derivative of any term with a #y# when implicitly differentiating will put the chain rule into effect and split out a #dy/dx# term. Also remember that finding #d/dx[ytanx]# will require the product rule.

#0=tanxdy/dx+yd/dx[tanx]-2ydy/dx#

#0=tanxdy/dx+ysec^2x-2ydy/dx#

#2ydy/dx-tanxdy/dx=ysec^2x#

#dy/dx(2y-tanx)=ysec^2x#

#dy/dx=(ysec^2x)/(2y-tanx)#