Question

How do you implicitly differentiate: sqrt(xy)=x^2y+1

Answer 1

a derivative is a derivative is a derivative; the rules of differentiation are the same no matter what your looking at

Answer 2

Is the answer: \[y'=(2x)/(2\sqrt(xy))\]

Answer 3

dunno, what steps did you take to come to that conclusion?

Answer 4

\[1/2(xy)^{-1/2}=2xy'\]
then
\[1/2\sqrt(xy)=2xy'\]

Answer 5

I used only product and power rule.

Answer 6

sqrt(xy) = x^2 y + 1

sqrt(xy)' = (x^2 y)' + 1'

sqrt(xy)' (xy)'= (x^2 y)' + 1'

(xy)'/2sqrt(xy) = x'^2 y + x^2 y'

(x'y+xy')/2sqrt(xy) = 2x x' y + x^2 y'

if we assume that this is diffed wrt x then x'=1

(y+xy')/2sqrt(xy) = 2xy + x^2 y'

Answer 7

(y+xy')/2sqrt(xy) = 2xy + x^2 y'

(y+xy')/2sqrt(xy) - x^2 y'= 2xy

y' (y+x)/2sqrt(xy) - x^2= 2xy

y' = 2xy/((y+x)/2sqrt(xy) - x^2)

\[ \large y' = \cfrac{2xy}{\frac{y+x}{2\sqrt{xy}} - x^2}\]