How do you find the derivative of #sqrtx+sqrty=9#?

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Answer 1 · 2017-04-13T02:29:48

#dy/dx=-sqrt(y/x)#

#sqrt(x)+sqrt(y)=9#

By rewriting a bit,

#Rightarrow x^(1/2)+y^(1/2)=9#

By differentiating with respect to #x#,

#Rightarrow 1/2x^(-1/2)+1/2 y^(-1/2)dy/dx=0#

By cleaning up a bit,

#Rightarrow 1/(2sqrt(x))+1/(2sqrt(y))dy/dx=0#

By subtracting #1/(2sqrt(x))# from both sides,

#1/(2sqrt(y))dy/dx=-1/(2sqrt(x))#

By multiplying both sides by #2sqrt(y)#,

#dy/dx=-(cancel(2)sqrt{y})/(cancel(2)sqrt(x))=-\sqrt(y/x)#

I hope that this was clear.