1. ## Calculus Implicit Differentiation

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

Thanks!

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

Thanks!
$\frac{d}{dx} xy = \frac{d}{dx} (2x^2-3y)$

$y + x\frac{dy}{dx} = 4x-3\frac{dy}{dx}$

$x\frac{dy}{dx}+3\frac{dy}{dx} = 4x-y$

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

$\frac{dy}{dx} = \frac{4x-y}{x+3}$

3. Hello, SoYah!

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

We have: . $-\sin(xy)\left(x\frac{dy}{dx} + y\right) \:=\:4x - 3\frac{dy}{dx}$

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

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

Factor: . $\bigg[3-x\sin(xy)\bigg]\frac{dy}{dx} \:=\:4x + y\sin(xy)$

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

4. Originally Posted by Mush
$\frac{d}{dx} xy = \frac{d}{dx} (2x^2-3y)$

$y + x\frac{dy}{dx} = 4x-3\frac{dy}{dx}$

$x\frac{dy}{dx}+3\frac{dy}{dx} = 4x-y$

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

$\frac{dy}{dx} = \frac{4x-y}{x+3}$
You missed the cos on the left hand side

5. Originally Posted by mr fantastic
You missed the cos on the left hand side
Ahhh, so I did.