How to Find the Equation of the Tangent Line to the Curve of $x^{12}y^{4}-x^9y^{10}=0$ at the Point (1,1)? I am given the following equation: $x^{12}y^{4}-x^9y^{10}=0$
I am to find the equation of the tangent line to the curve of the equation above at the point (1,1).
So far, I have made the following steps:
1) Using the product rule, find the derivative of the function
2) Arrive at the following derivative: $x^{12}4y^3+12x^{11}y^4-x^910y^9-9x^8y^{10}$
3) Sub in the value of 1 for all x variables in the equation of step 2.
4) Arrive at $4y^3+12y^4-10y^9-9y^{10}$
5) This is the step where I am not sure what to do. I believe I am to get 'y' onto one side, in order to get the slope of the tangent line, but I am not sure how.
Does anyone know where I should go from here?
All help is appreciated.
 A: My observation is that your curve is $$x^9y^4\left(x^3-y^6\right)=0$$
So it is the superposition of the $y$-axis, the $x$-axis, and $y^6=x^3$. This last part is where your point lives. And this last part is the same as $\left\lvert y\right\rvert=\sqrt{x}$. And on the portion where $(1,1)$ lives, it's the same as $y=\sqrt{x}$. 
So if implicit differentiation is unfamiliar to you, you can just use direct differentiation: $\frac{dy}{dx}=\frac{1}{2\sqrt{x}}$. And the slope at this point is $\frac{1}{2}$. And so the line is $y-1=\frac{1}{2}(x-1)$.

But here is the problem with what you did. In 1), you say "find the derivative of the function". I think that is misuse of language. A function is a device for converting inputs into outputs. That is not the purpose of the expression you have here. You have an expression. And whenever $x$ and $y$ are such that the expression results in $0$, you have a point on a curve. If there is any function here, it is $y$ as a function of $x$. Because if you are at a point like $(1,1)$, then changing $x$ a little will induce a change in $y$ for you to stay on that curve. So in that sense, $y$ depends on $x$.
To get from 1) to 2), try thinking of it as applying $\frac{d}{dx}$ to both sides. Of course $\frac{d}{dx}x=1$, but $\frac{d}{dx}y=\frac{dy}{dx}$, and does not simplify further. So you should have $$12x^{11}y^4+4x^{12}y^3\frac{dy}{dx}-9x^8y^{10}-10x^9y^9\frac{dy}{dx}=0$$ Now you can let $x=y=1$: $$12+4\left.\frac{dy}{dx}\right|_{(1,1)}-9-10\left.\frac{dy}{dx}\right|_{(1,1)}=0$$ and take it from here to find $\left.\frac{dy}{dx}\right|_{(1,1)}$. And with that, use the point-slope form of a line equation to get the tangent line's equation.
A: This way is more elegant:
Let $C := \{ (x,y) \in \Bbb{R}^{2} \mid x^{12}y^{4} - x^{9}y^{10} = 0 \}$; let $f: (x,y) \mapsto x^{12}y^{4} - x^{9}y^{10}: \Bbb{R}^{2} \to \Bbb{R}$. Then
$\nabla f(x,y) = (12x^{11}y^{4} - 9x^{8}y^{10}, 4x^{12}y^{3} - 10x^{9}y^{9})$ for all $(x,y) \in \Bbb{R}^{2}$, implying that 
$\nabla f(1,1) = (3,-6)$. But $\nabla f(1,1)$ is normal to $C$ at $(1,1)$; hence the tangent line to $C$ through $(1,1)$ is the set of all points $(x,y) \in \Bbb{R}^{2}$ such that $(x-1, y-1)\cdot (3,-6) = 0$.
A: You should use implicit differentiation to arrive at the derivative.
$$4 x^{12} y^3 \frac{dy}{dx}+12 x^{11} y^4-10 x^9 y^9 \frac{dy}{dx}-9 x^8 y^{10}=0 \\ 
\Rightarrow \left(4 x^{12} y^3-10 x^9 y^9\right) \frac{dy}{dx}=9 x^8 y^{10}-12 x^{11} y^4 \\
\Rightarrow \frac{dy}{dx} ={(-12 x^{11} y^4+9 x^8 y^{10})\over(4 x^{12} y^3-10 x^9 y^9)}$$
Now you can use the point $(1,1)$ to find the slope of the tangent line at that point.
A: Using step 1, you need to define the variable you are taking the derivative of.  The point is you can do the same thing to both sides of the equation.  You have taken the derivative with respect to $x$.  You have lost the derivative of $y$ with respect to $x$, so you should get $x^{12}4y^3y'+12x^{11}y^4-x^910y^9y'-9x^8y^{10}=0$ or $(x^{12}4y^3-10x^9y^9)y'-12x^{11}y^4-9x^8y^{10}=0$  Now you can evaluate $y'$ at $(1,1)$
A: Differentiate implicitly to get: $$12x^{11}y^4+4x^{12}y^3y'-9x^8y^{10}-10x^9y^9y'=0$$
You can factor out the $y'$ and do some algebra to get the rest.
