How do you find #dy/dx# by implicit differentiation of #x^(1/2)+y^(1/2)=9#?

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Answer 1 · 2017-01-11T10:32:45

#(dy)/(dx) = 1-9x^(-1/2)#

Differentiate the equation with respect to #x#, considering that:

#d/(dx) f(y(x)) = (df)/(dy)*y'(x)#

so:

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

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

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

From the original equation we see that:

#y^(1/2) = 9-x^(1/2)#

and we have:

#(dy)/(dx) = x^(-1/2)(x^(1/2)-9) = 1-9x^(-1/2)#