Think y {\displaystyle y} as a function of x {\displaystyle x} .
x y = x 2 y − 2 {\displaystyle {\sqrt {xy}}=x^{2}y-2} , Because these are equal, the derivatives must be equal:
d ( x y ) d x = d ( x 2 y ) d x − d ( − 2 ) d x 1 2 ⋅ ( x y ) − 1 / 2 ⋅ ( y + x d y d x ) (using chain rule and product rule) = 2 x y + x 2 d y d x (product rule) 1 2 y x + 1 2 x y d y d x = 2 x y + x 2 d y d x 1 2 y x − 2 x y = − 1 2 x y d y d x + x 2 d y d x 1 2 y x − 2 x y = ( − 1 2 x y + x 2 ) d y d x d y d x = 1 2 y x − 2 x y − 1 2 x y + x 2 {\displaystyle {\begin{aligned}&{\frac {d({\sqrt {xy}})}{dx}}={\frac {d(x^{2}y)}{dx}}-{\frac {d(-2)}{dx}}\\&{\frac {1}{2}}\cdot (xy)^{-1/2}\cdot (y+x{\frac {dy}{dx}}){\text{(using chain rule and product rule)}}=2xy+x^{2}{\frac {dy}{dx}}{\text{(product rule)}}\\&{\frac {1}{2}}{\sqrt {\frac {y}{x}}}+{\frac {1}{2}}{\sqrt {\frac {x}{y}}}{\frac {dy}{dx}}=2xy+x^{2}{\frac {dy}{dx}}\\&{\frac {1}{2}}{\sqrt {\frac {y}{x}}}-2xy=-{\frac {1}{2}}{\sqrt {\frac {x}{y}}}{\frac {dy}{dx}}+x^{2}{\frac {dy}{dx}}\\&{\frac {1}{2}}{\sqrt {\frac {y}{x}}}-2xy=(-{\frac {1}{2}}{\sqrt {\frac {x}{y}}}+x^{2}){\frac {dy}{dx}}\\&{\frac {dy}{dx}}={\frac {{\frac {1}{2}}{\sqrt {\frac {y}{x}}}-2xy}{-{\frac {1}{2}}{\sqrt {\frac {x}{y}}}+x^{2}}}\end{aligned}}}
Therefore, the derivative of y {\displaystyle y} at point ( 1 , 4 ) {\displaystyle (1,4)} is
1 2 y x − 2 x y − 1 2 x y + x 2 = 1 2 4 1 − 8 − 1 2 1 4 + 1 2 = − 28 3 {\displaystyle {\frac {{\frac {1}{2}}{\sqrt {\frac {y}{x}}}-2xy}{-{\frac {1}{2}}{\sqrt {\frac {x}{y}}}+x^{2}}}={\frac {{\frac {1}{2}}{\sqrt {\frac {4}{1}}}-8}{-{\frac {1}{2}}{\sqrt {\frac {1}{4}}}+1^{2}}}={\frac {-28}{3}}}
answer: − 28 3 {\displaystyle \color {blue}{\frac {-28}{3}}}
as a function of 
.
, Because these are equal, the derivatives must be equal:
at point 
is
