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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.

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1 Answer 1

up vote 2 down vote accepted

Very well done, Fernando. You indeed found the equation for the line tangent to the second equation at the point $(2, 1)$, and your first solution look like you differentiated properly, too.

You could simplify just a tad: $$\frac{dy}{dx}=\frac{2x-4y^3-16}{-4(x)(3y^2)+8x^2} = \frac{2(x-2y^3 - 8)}{4(-3xy^2 +2)} = \frac{x - 2y^3 - 8}{2(-3xy^2 + 2)}$$

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thanks very much. –  Fernando Martinez Feb 21 '13 at 18:32
    
You're welcome, Fernando! –  amWhy Feb 21 '13 at 18:32
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