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:

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

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