How do you implicitly differentiate $y/x^4-3/y^2-y=8$ ?

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Answer 1 · 2017-12-01T08:44:08

$(dy)/(dx)=(4yx^(-5))/(x^(-4)+6y^(-3)-1)$

First, the equation will be rewritten to get $yx^(-4)-3y^(-2)-y=8$

Now we differentiate with respect to $x$ :

$d/(dx)[yx^(-4)-3y^(-2)-y]=d/(dx)[8]$

$d/(dx)[yx^(-4)]-d/(dx)[3y^(-2)]-d/(dx)[y]=d/(dx)[8]$

$d/(dx)[y]x^(-4)-4yx^(-5)-d/(dx)[3y^(-2)]-d/(dx)[y]=0$

Using the chain rule, we get $d/(dx)=d/(dy)*(dy)/(dx)$

$d/(dy)[y]* (dy)/(dx)x^(-4)-4yx^(-5)-d/(dy)[3y^(-2)]* (dy)/(dx)-d/(dy)[y]*(dy)/(dx)=0$

$(dy)/(dx)x^(-4)-4yx^(-5)+6y^(-3)* (dy)/(dx)-(dy)/(dx)=0$

$(dy)/(dx)[x^(-4)+6y^(-3)-1]-4yx^(-5)=0$

$(dy)/(dx)[x^(-4)+6y^(-3)-1]=4yx^(-5)$

$(dy)/(dx)=(4yx^(-5))/(x^(-4)+6y^(-3)-1)$

How do you implicitly differentiate y/x^4-3/y^2-y=8 y/x^4-3/y^2-y=8 ?