# Calculus Implicit Differentiation

• Jan 28th 2009, 05:04 PM
SoYah
Calculus Implicit Differentiation
Use implicit differentiation to find dy/dx if cos xy = 2x^2-3y

Thanks!
• Jan 28th 2009, 05:46 PM
Mush
Quote:

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

Thanks!

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

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

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

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

$\displaystyle \frac{dy}{dx} = \frac{4x-y}{x+3}$
• Jan 28th 2009, 06:20 PM
Soroban
Hello, SoYah!

Quote:

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

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

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

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

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

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

• Jan 28th 2009, 06:28 PM
mr fantastic
Quote:

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

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

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

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

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

You missed the cos on the left hand side (Doh)
• Jan 28th 2009, 06:30 PM
Mush
Quote:

Originally Posted by mr fantastic
You missed the cos on the left hand side (Doh)

Ahhh, so I did.