# Thread: differentiating in terms of y and x

1. ## differentiating in terms of y and x

Given that
$\displaystyle siny=xy+x^2$
find dy/dx in terms of $\displaystyle x$ and $\displaystyle y$

Plese talk me through how to do this.
Thanks

2. Originally Posted by George321
Given that
$\displaystyle siny=xy+x^2$
find dy/dx in terms of $\displaystyle x$ and $\displaystyle y$

Plese talk me through how to do this.
Thanks
Have you been taught implicit differentiation? There are many threads in this subforum that cover questions like this one. Use the Search tool to find them, read them and then make another attempt on this question. You should also review the examples in your class notes and textbook.

Please post your work and say where you are stuck if you need more help after doing this.

Here is a small start:

From the product rule:

$\displaystyle \frac{d (xy)}{dx} = y + x \frac{dy}{dx}$.

From the chain rule:

$\displaystyle \frac{d (\sin y)}{dx} = \frac{d (\sin y)}{dy} \cdot \frac{dy}{dx}$.

3. so...
$\displaystyle siny-xy=x^2$

$\displaystyle 2x=cosy\frac{dy}{dx}-x\frac{dy}{dx}-y$

$\displaystyle 2x+y=\frac{dy}{dx}(cosy-x)$

$\displaystyle \frac{dy}{dx}=\frac{2x-y}{cosy-x}$

Is that correct?

4. Originally Posted by George321
so...
$\displaystyle siny-xy=x^2$

$\displaystyle 2x=cosy\frac{dy}{dx}-x\frac{dy}{dx}-y$

$\displaystyle 2x+y=\frac{dy}{dx}(cosy-x)$

$\displaystyle \frac{dy}{dx}=\frac{2x-y}{cosy-x}$

Is that correct?
Yes, except for the typo you made (should be $\displaystyle 2x+y$ in the numerator of the final answer).