Answer to Question #90995 in Calculus for Rachel

Question

Differentiate the following:
5ln(2x−2y)−2y^2+5x=0

Answer 1

$$F=5ln(2x-2y)-2y^2+5x$$

In calculus, a method called implicit differentiation makes use of the chain rule to differentiate implicitly defined functions

$$y'=-\frac{F'_x}{F'_y}$$

take the derivative with respect to x from F

$${F'_x}=5ln(2x-2y)dx-2y^2dx+5xdx$$

$$(5ln(2x-2y))'dx=5(ln(2x-2y)'dx$$

$$(u(v))'=u'(v)v'$$

$$5(ln(2x-2y)'dx=5\frac{(2x-2y)'dx}{2x-2y}=5\frac{2}{2x-2y}$$

$$(2y^2)'dx=0$$

$$(5x)'dx=5$$

$${F'_x}==5\frac{2}{2x-2y}+5$$

take the derivative with respect to y from F

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$${F'_y}=5ln(2x-2y)dy-2y^2dy+5xdy$$

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$$(5ln(2x-2y)'dy=5(ln(2x-2y)'dy=5\frac{(2x-2y)'dy}{2x-2y}=5\frac{-2}{2x-2y}$$

$$(2y^2)'dy=4y$$

$$(5x)'dy=0$$

$${F'_y}==5\frac{-2}{2x-2y}-4y$$

$${F'_x}=\frac{5}{x-y}+5=5\frac{1+x-y}{x-y}$$

$${F'_y}=-(\frac{5}{x-y}+4y)=-(\frac{5+4y(x-y)}{x-y})$$

$$y'=-\frac{5\frac{1+x-y}{x-y}}{-(\frac{5+4y(x-y)}{x-y})}$$

$$y'=\frac{5\frac{1+x-y}{x-y}}{\frac{5+4y(x-y)}{x-y}}$$ $$y'=\frac{5(1+x-y)}{5+4y(x-y)}$$