What is the implicit derivative of #1= ysqrt(xy)-y #?

Question

What is the implicit derivative of #1=  ysqrt(xy)-y  #?

Evelyn Agard · Accepted Answer

#dy/dx=-(ysqrt(xy))/(x(3sqrt(xy)-2))#

#d/dx[1=ysqrt(xy)-y]#

#0=sqrt(xy)dy/dx+yd/dx[sqrt(xy)]-dy/dx#

#0=sqrt(xy)dy/dx+y(1/(2sqrt(xy)))d/dx[xy]-dy/dx#

#0=sqrt(xy)dy/dx+y/(2sqrt(xy))(y+xdy/dx)-dy/dx#

#0=sqrt(xy)dy/dx+y^2/(2sqrt(xy))+(xy)/(2sqrt(xy))(dy/dx)-dy/dx#

#0=sqrt(xy)dy/dx+(ysqrt(xy))/(2x)+(sqrt(xy))/2(dy/dx)-dy/dx#

#-(ysqrt(xy))/(2x)=dy/dx(sqrt(xy)+(sqrt(xy))/2-1)#

#-(ysqrt(xy))/(2x)=dy/dx((3sqrt(xy)-2)/2)#

#dy/dx=-(ysqrt(xy))/(2x)(2/(3sqrt(xy)-2))#

#dy/dx=-(ysqrt(xy))/(x(3sqrt(xy)-2))#

Please ask with any questions! Above all, never forget to use the product rule.

Luke Alvarez · Answer

undefined To find the implicit derivative of the equation $1 = y \sqrt{xy} - y$, differentiate both sides of the equation with respect to $x$, treating $y$ as a function of $x$ using the chain rule and product rule where necessary.

$1 = y \sqrt{xy} - y$

Differentiating both sides with respect to $x$:

$0 = y' \sqrt{xy} + y \frac{1}{2\sqrt{xy}}(x) + y'\sqrt{xy} - y'$

Simplify and solve for $y'$:

$0 = 2y' \sqrt{xy} - y' + \frac{y}{2\sqrt{xy}}$

$y' = \frac{y}{2\sqrt{xy}} / (2\sqrt{xy} - 1)$