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
$\displaystyle \frac{dy}{dx}=\cos(xy)\cdot \frac{d}{dx}(xy)$
Now we need to use the product rule to finish taking the derivative
$\displaystyle \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
$\displaystyle \frac{dy}{dx}-x\cos(xy)\frac{dy}{dx}=y\cos(xy)$
$\displaystyle \frac{dy}{dx}(1-x\cos(xy))=y\cos(xy)$
$\displaystyle \frac{dy}{dx}=\frac{y\cos(xy)}{1-x\cos(xy)}$
The implict derivative of
$\displaystyle \frac{d}{dx}(xy)$
uses both the product rule and chain rule
First the product rule give
$\displaystyle \frac{d}{dx}(dx)=\left( \frac{d}{dx}x\right) \cdot y + x\left( \frac{d}{dx} y\right)$
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 $\displaystyle \frac{d}{dx}y=\frac{dy}{dx}=y'$
$\displaystyle \frac{d}{dx}(dx)=\left( \frac{d}{dx}x\right) \cdot y + x\left( \frac{d}{dx} y\right)=1\cdot y+x\cdot \frac{dy}{dx}=y+xy'$
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