# Thread: find dy/dx by implicit differentiation

1. ## find dy/dx by implicit differentiation

Can someone show me step by step how to do

y= sin xy

2. Originally Posted by Nightasylum
Can someone show me step by step how to do

y= sin xy
$y = \sin(xy)$

$y' = \cos(xy) \cdot (xy' + y)$

$y' = xy'\cos(xy) + y\cos(xy)$

$y' - xy'\cos(xy) = y\cos(xy)$

$y'[1 - x\cos(xy)] = y\cos(xy)$

$y' = \frac{y\cos(xy)}{1 - x\cos(xy)}$

3. Originally Posted by Nightasylum
Can someone show me step by step how to do

y= sin xy
first we need to use the chain rule on the right hand side

$\frac{dy}{dx}=\cos(xy)\cdot \frac{d}{dx}(xy)$

Now we need to use the product rule to finish taking the derivative

$\frac{dy}{dx}=\cos(xy)\cdot \left( 1 \cdot y+x\frac{dy}{dx} \right)=y\cos(xy)+x\cos(xy)\frac{dy}{dx}$

Now we just need to solve for the derivative

$\frac{dy}{dx}-x\cos(xy)\frac{dy}{dx}=y\cos(xy)$

$\frac{dy}{dx}(1-x\cos(xy))=y\cos(xy)$

$\frac{dy}{dx}=\frac{y\cos(xy)}{1-x\cos(xy)}$

4. Im still not grasping how to do this , I see you get the derivative of sin times the stuff in the ( ) right? but where does the +Y come from

5. Originally Posted by Nightasylum
Im still not grasping how to do this , I see you get the derivative of sin times the stuff in the ( ) right? but where does the +Y come from
it's the product rule ...

the derivative of $(xy)$ w/r to $x$ is ...

$x \cdot y' + y\cdot 1$

6. Originally Posted by Nightasylum
Im still not grasping how to do this , I see you get the derivative of sin times the stuff in the ( ) right? but where does the +Y come from

The implict derivative of

$\frac{d}{dx}(xy)$

uses both the product rule and chain rule

First the product rule give

$\frac{d}{dx}(dx)=\left( \frac{d}{dx}x\right) \cdot y + x\left( \frac{d}{dx} y\right)$

i.e the dervitive of the first times the 2nd plus the derviatve of the 2nd times the first

Now using implict differenation we get and

Remembering that $\frac{d}{dx}y=\frac{dy}{dx}=y'$

$\frac{d}{dx}(dx)=\left( \frac{d}{dx}x\right) \cdot y + x\left( \frac{d}{dx} y\right)=1\cdot y+x\cdot \frac{dy}{dx}=y+xy'$

7. Thanks guys I see what im doing now wish I had some tutor here that was as helpful as you all. Last time I made a stop by the tutor on campus they had no clue what I was doing and my professor only allows 30 mins during their office hours because she trys to help everyone. /sigh