Question

find an equation for the tangent to the curve given by y - sin (sqrt (xy)) = pi at the point (pi, pi)

Answer 1

dy/dx - 0.5(1/sqrt(xy))*(y + (xdy)/dx)*cos(sqrt(xy)) = 0

dydx = ycos(sqrt(xy) / (2sqrt(xy) - xcos(sqrt(xy)))

plug x = pi , y = pi

-pi / (2pi + pi) = -pi / 3pi = -1/3

Answer 2

is it ok ?

Answer 3

\[y-\sin{\sqrt{xy}}=\pi\quad/\frac{d}{dy}\]\[1-\frac{\cos{\sqrt{xy}}}{2\sqrt{xy}}(x'y+x)=0\]\[x'=\frac{1}{y}\left(\frac{2\sqrt{xy}}{\cos{\sqrt{xy}}}-x\right)\]\[x'(\pi,\pi)=\frac{1}{\pi}\left(\frac{2\pi}{\cos{\pi}}-\pi\right)=-3\]\[y'=\frac{1}{x}=-\frac{1}{3}\]tangent line:\[y-\pi=-\frac{1}{3}(x-\pi)\] Coolsector this is OK

Answer 4

\[y'=\frac{1}{x'}\]

Answer 5

yes forgot about finding the line.. thought it was all about the slope (which it is anyway lol)