<s>[INST] Implicit differentiation 2 questions? Hello everyone I have two questions on implicit differentiation. 
My first one is express $\frac{dy}{dx}$ in terms of $x$ and $y$ if $x^2-4xy^3+8x^2y=20$
what I did is this 
$2x-4x(3y^2)(\frac{dy}{dx})+4y^3+8x^2\frac{dy}{dx}+16xy=0$
$-4x(3y^2\frac{dy}{dx})+8x^2\frac{dy}{dx}=-2x-4y^3-16$
Finally I got $\frac{dy}{dx}=\frac{2x-4y^3-16}{-4(x)(3y^2)+8x^2}$
My second question is 
Find an equation for the tangent line to the graph of $2x^2-5y^2+xy=5$ at the point $(2,1)$
Anyway I simplified the equation to 
$\frac{dy}{dx}=\frac{-4x-1}{-10y+x}$ plugging in x and y I got $\frac{9}{8}$
so my line is $y-1=\frac{9}{8}(x-2)$ but I am unsure if if I did this correctly.
 [/INST] There are slight computation errors and omission errors; Other than that you are doing fine. If you experience such errors, it is better to write every step.
In the first one,
when you differentiate both sides you get $2x-4(x3y^2 dy/dx +y^3)+8(2xy+x^2 dy/dx)=0$. Thus $(-4x3y^2+8x^2)dy/dx=-2x+4y^3-16xy$ so that $dy/dx=\frac{-2x+4y^3-16xy}{-12xy^2+8x^2}=\frac{-x+2y^3-8xy}{-6xy^2+4x^2}$
In the second one, a similar error is there.
Differentiating gives $4x-10y dy/dx+(x dy/dx +y)=0$ and thus $(-10y+x)dy/dx=-4x+y$ so that $dy/dx=\frac{-4x-y}{-10y+x}$. The value you get after substitution is accidentally(?) same here.
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