Frustrated by implicit differentiation problem

Question

I've been working on the following problem for over an hour. I think I'm making a stupid mistake. I'm supposed to find equation of the tangent line to the curve at point (3,1) for the following function:

$ 3\left(x^2+y^2\right)^2\:=\:100xy^2 $

Getting $ \frac{dy}{dx} $ for the LHS gets me 360 + 336$ \frac{dy}{dx} $ after I differentiate and plug in x and y above. But differentiating the RHS is much more confusing for me...I think it's $ 100y^2+200xy\:\frac{dy}{dx} $? I think? I'm not sure, I ended up kind of guessing based on notes provided by the homework. Solving for $ \frac{dy}{dx} $ got me $\frac{65}{66}$. Shouldn't this mean the equation for the point at (3,1) be $y\:=\:\frac{65}{66}x-\frac{43}{22}$? According to the homework, this is incorrect (I'm doing hw online). I feel like I'm either making a stupid mistake in my algebra, or that I incorrectly differentiated the LHS.

Answer 1

Hint:

Your RHS is correct. Differentiating both sides (and letting $y=y(x)$), we have

$$6\left(x^2+y^2\right)\left(2x+2y\frac{\mathrm{d}y}{\mathrm{d}x}\right)=100\left(y^2+2xy \frac{\mathrm{d}y}{\mathrm{d}x}\right).$$

Now insert your point $(3,1)$ and isolate $ \frac{\mathrm{d}y}{\mathrm{d}x}$ (you should get $\frac{\mathrm{d}y}{\mathrm{d}x}=\frac{13}{24}$). Let me know if you need more help than this.

Edit in response to comment (LHS differentiation):

$$3\frac{\mathrm{d}}{\mathrm{d}x}\left(x^2+y^2\right)^2=6\left(x^2+y^2\right)\frac{\mathrm{d}}{\mathrm{d}x}\left(x^2+y^2\right)=6\left(x^2+y^2\right)\left(2x+\frac{\mathrm{d }(y^2)}{\mathrm{d}x}\right)=6\left(x^2+y^2\right)\left(2x+2y\frac{\mathrm{d }y}{\mathrm{d}x}\right).$$

Answer 2

$\frac{d}{dx}[(x^2+y^2)^2] = 2(x^2+y^2)\frac{d}{dx}(x^2+y^2)\\\frac{d}{dx}(xy^2) = y^2 + x\frac{d}{dx}(y^2)$

Answer 3

$$3((3+dx)^2+(1+dy)^2)^2=100(3+dx)(1+dy)^2$$ $$3(9+6dx+1+2dy)^2=(300+100dx)(1+2dy)$$ $$3(100+120dx+40dy)=300+100dx+600dy$$ $$300+360dx+120dy=300+100dx+600dy$$ $$260dx=480dy$$ $$dy/dx=13/24$$