How do you implicitly differentiate $- y^{2} = e^{2 x - 4 y} - 2 \frac{y}{x}$ $-y^2=e^(2x-4y)-2y/x$ ?

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Answer 1 · 2016-05-28T07:08:33

$y'=(x^2*e^(2x-4y)+y)/(2x^2*e^(2x-4y) +x-x^2y)$

Start differentiating with respect to x both sides of the equation
$-y^2=e^(2x-4y) -2*y/x$

$d/dx(-y^2)=d/dx(e^(2x-4y)) -d/dx(2*y/x)$

$(-2y*y')=(e^(2x-4y)*d/dx(2x-4y)) -(2*d/dx(y/x))$

$-2y*y'=e^(2x-4y)(2-4y') -2*((xy'-y*1)/x^2)$

Simplify by dividing by 2

$-y*y'=e^(2x-4y)(1-2y') -((xy'-y*1)/x^2)$

$-y*y'=e^(2x-4y)-2*e^(2x-4y)y' -(xy')/x^2+(y)/x^2$

$-y*y'=e^(2x-4y)-2*e^(2x-4y)y' -(y')/x+y/x^2$

$2*e^(2x-4y)y' +(y')/x-y*y'=e^(2x-4y)+y/x^2$

Factor the y'

$(2*e^(2x-4y) +1/x-y)y'=e^(2x-4y)+y/x^2$

Divide both sides by $(2*e^(2x-4y) +1/x-y)$

$((2*e^(2x-4y) +1/x-y)y')/((2*e^(2x-4y) +1/x-y))=(e^(2x-4y)+y/x^2)/(2*e^(2x-4y) +1/x-y)$

$cancel((2*e^(2x-4y) +1/x-y)y')/cancel((2*e^(2x-4y) +1/x-y))=(e^(2x-4y)+y/x^2)/(2*e^(2x-4y) +1/x-y)$

$y'=(e^(2x-4y)+y/x^2)/(2*e^(2x-4y) +1/x-y)$

Simplify by multiplying the right side by $x^2/x^2$

$y'=((e^(2x-4y)+y/x^2))/((2*e^(2x-4y) +1/x-y))*x^2/x^2$

$y'=(x^2*e^(2x-4y)+y)/(2x^2*e^(2x-4y) +x-x^2y)$

God bless....I hope the explanation is useful.

How do you implicitly differentiate -y^2=e^(2x-4y)-2y/x −y2=e2x−4y−2yx-y^2=e^(2x-4y)-2y/x ?