How do you determine dy/dx given #ysqrtx+xsqrty=16#?

1 Answer
Oct 7, 2016

#y'=(-y/(2sqrtx)-sqrty)/(sqrtx+(x)/(2sqrty))#

Explanation:

We have to use implicit differentiation which is basically just a special case of the chain rule. Every time we differentiate a factor that has a #y# variable we have to include the factor #dy/dx# or #y'#.

We will also be using the product rule and power rule to answer this question.

Begin by changing all of the radicals to their exponent equivalents.

#yx^(1/2)+xy^(1/2)=16#

#y(1/2)x^(-1/2)+(1)y'x^(1/2)+x(1/2)y^(-1/2)y'+(1)y^(1/2)=0#

Simplify and switch back to radicals

#y/(2sqrtx)+y'sqrtx+(xy')/(2sqrty)+sqrty=0#

Gather terms with #y'# to one side and everything else to the other side

#y'sqrtx+(xy')/(2sqrty)=-y/(2sqrtx)-sqrty#

Factor out #y'#

#y'(sqrtx+(x)/(2sqrty))=-y/(2sqrtx)-sqrty#

Isolate #y'# by dividing

#y'cancel(sqrtx+(x)/(2sqrty))/cancel(sqrtx+(x)/(2sqrty))=(-y/(2sqrtx)-sqrty)/(sqrtx+(x)/(2sqrty))#

#y'=(-y/(2sqrtx)-sqrty)/(sqrtx+(x)/(2sqrty))#

Click on the tutorial to see another example.