Find by `dy/dx` implicit differentiation `x^2-4xy+y^2=4`

Wednesday, November 10, 2010

Find by $\displaystyle\frac{{\left.{d}{y}}}{{\left.{d}{x}}}$ implicit differentiation $\displaystyle{{x}}^{{2}}-{4}{x}{y}+{{y}}^{{2}}={4}$

We are asked to find $\displaystyle\frac{{{\left.{d}{y}\right.}}}{{{\left.{d}{x}\right.}}}$ if $\displaystyle{{x}}^{{2}}-{4}{x}{y}+{{y}}^{{2}}={4}$

Note that this is difficult to write as a function of x, so we take the derivative implicitly:

Working term by term:

$\displaystyle\frac{{d}}{{{\left.{d}{x}\right.}}}{\left({{x}}^{{2}}\right)}={2}{x}$

$\displaystyle\frac{{d}}{{{\left.{d}{x}\right.}}}{\left(-{4}{x}{y}\right)}=-{4}{y}-{4}{x}\frac{{{\left.{d}{y}\right.}}}{{{\left.{d}{x}\right.}}}$ using the product rule

$\displaystyle\frac{{d}}{{{\left.{d}{x}\right.}}}{\left({{y}}^{{2}}\right)}={2}{y}\frac{{{\left.{d}{y}\right.}}}{{{\left.{d}{x}\right.}}}$

So we get $\displaystyle{2}{x}-{4}{y}-{4}{x}\frac{{{\left.{d}{y}\right.}}}{{{\left.{d}{x}\right.}}}+{2}{y}\frac{{{\left.{d}{y}\right.}}}{{{\left.{d}{x}\right.}}}={0}$

$\displaystyle\frac{{{\left.{d}{y}\right.}}}{{{\left.{d}{x}\right.}}}{\left({2}{y}-{4}{x}\right)}={4}{y}-{2}{x}$

$\displaystyle\frac{{{\left.{d}{y}\right.}}}{{{\left.{d}{x}\right.}}}=\frac{{{4}{y}-{2}{x}}}{{{2}{y}-{4}{x}}}=\frac{{{2}{y}-{x}}}{{{y}-{2}{x}}}$

-------------------------------------------------------

$\displaystyle\frac{{{\left.{d}{y}\right.}}}{{{\left.{d}{x}\right.}}}=\frac{{{x}-{2}{y}}}{{{2}{x}-{y}}}$

Ninenotes

Wednesday, November 10, 2010

Find by $\displaystyle\frac{{\left.{d}{y}}}{{\left.{d}{x}}}$ implicit differentiation $\displaystyle{{x}}^{{2}}-{4}{x}{y}+{{y}}^{{2}}={4}$

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find square roots of -1+2i

Wednesday, November 10, 2010

Find by implicit differentiation

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find square roots of -1+2i

Find by $\displaystyle\frac{{\left.{d}{y}}}{{\left.{d}{x}}}$ implicit differentiation $\displaystyle{{x}}^{{2}}-{4}{x}{y}+{{y}}^{{2}}={4}$