Math Help - weird implicit differentiation problem

1. weird implicit differentiation problem

Use implicit differentiation to find dy/dx if cos xy = 2x² -3y.

My steps :

(-sin(xy))(x(dy/dx) + y) = 4x - 3(dy/dx)

(-sin(xy))(x(dy/dx) + y) + 3(dy/dx) = 4x

-xsin(xy)(dy/dx) - ysin(xy) + 3(dy/dx) = 4x

-xsin(xy)(dy/dx) + 3(dy/dx) = 4x + ysin(xy)

dy/dx (3 - xsin(xy)) = (4x +ysin(xy))

dy/dx = (4x +ysin(xy)) / (3 - xsin(xy))

Are my steps correct? I am not sure of the answer, cuz it looks super weird.

2. Originally Posted by h4hv4hd4si4n
Use implicit differentiation to find dy/dx if cos xy = 2x² -3y.

My steps :

(-sin(xy))(x(dy/dx) + y) = 4x - 3(dy/dx)

(-sin(xy))(x(dy/dx) + y) + 3(dy/dx) = 4x

-xsin(xy)(dy/dx) - ysin(xy) + 3(dy/dx) = 4x

-xsin(xy)(dy/dx) + 3(dy/dx) = 4x + ysin(xy)

dy/dx (3 - xsin(xy)) = (4x +ysin(xy))

dy/dx = (4x +ysin(xy)) / (3 - xsin(xy))

Are my steps correct? I am not sure of the answer, cuz it looks super weird.
You're correct!!

These implicit differentiation problems tend to have solutions that look like this (and sometimes worse! ).

3. I got the same answer, but I really need to refresh my memory on this topic. Thanks for the review. :P

5. Hello, h4hv4hd4si4n!

Use implicit differentiation to find $\tfrac{dy}{dx}\text{ if }\cos(xy) \:=\: 2x^2 -3y$

My steps:

$-\sin(xy)\,\left(x\tfrac{dy}{dx} + y\right) \;= \;4x - 3\tfrac{dy}{dx}$

$-x\sin(xy)\tfrac{dy}{dx} - y\sin(xy) \;=\; 4x - 3\tfrac{dy}{dx}$

$-x\sin(xy)\tfrac{dy}{dx} + 3\tfrac{dy}{dx} \;=\; 4x + y\sin(xy)$

$\tfrac{dy}{dx}\left[3 - x\sin(xy)\right] \;=\;4x +y\sin(xy)$

. . . $\frac{dy}{dx} \;=\; \frac{4x +y\sin(xy)}{3 - x\sin(xy)}$

Are my steps correct?

Absolutely! . . . Lovely work!