Find $y'$ for the equation: $xe^{-y/2} + ye^{-x/2} = \sin(x^2+y^2)$

Question

Find $y'$ for the equation: $$ xe^{-y/2} + ye^{-x/2} = \sin(x^2+y^2) $$

I have tried all ways I know, but I always have the problem of not noticing the important point. And end up going beyond in trying different methods.

We took and I tried the Implicit differentiation and the derivative of the logarithmic functions and I have been trying using those. (Note after answer. Apparently, all implicit functions we tried before, you would be able to replace the y's. Not this one though)

I have tried taking the derivative in multiple points of using these ways to get some numbers that I can work out but I didn’t get any thing useful.

I tried using the inverse of sin or expanding it.

I have reached equations like: $$ \frac{y+x}{2} + \ln\left(\frac{\sin(x^2+y^2)}{x\sqrt{e^x}+y\sqrt{e^y}}\right) = 0 $$ or $$ \frac{1-xy’}{\sqrt{e^x}} + \frac{y’-y/2}{\sqrt{e^y}} = \cos(x^2+y^2) (2x+2yy’) $$ and couple other.

Answer 1

Using your knowledge of the chain rule and the product rule, break the implicit function into parts. Thus:

$${d\over dx}xe^{-y/2}=x\cdot -{1\over 2}y'e^{-y/2} + e^{-y/2}$$

Then:

$${d\over dx}ye^{-x/2}=y\cdot -{1\over 2}e^{-x/2} + y'\cdot e^{-x/2}$$

Finally, the other side:

$${d\over dx}\sin(x^2+y^2)=(2x+2yy')\cos(x^2+y^2)$$

Combining everything together:

$$-{1\over 2}xy'e^{-y/2} + e^{-y/2} -{1\over 2}ye^{-x/2} + y'e^{-x/2} = (2x+2yy')\cos(x^2+y^2)$$

Then solve for $y'$.

Answer 2

I don't really have much idea how you got either of your equations, and they both appear to be rather wrong. The point of the exercise is to use the chain rule and implicitly differentiate. For example, the derivative of the first term on the left is

\begin{align} \frac{d}{dx} x e^{-y/2} &= e^{-y/2} \frac d {dx} x + x \frac{d}{dx} e^{-y/2} \\ &= e^{-y/2} + x e^{-y/2} \frac d {dx} \left(-\frac y2\right) \\ &= e^{-y/2} - \frac 1 2 xe^{-y/2} y'. \end{align}

Treat the other two parts similarly, and you'll be left with a linear equation to solve for $y'$.