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

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Answer 1 · 2016-01-30T10:29:27

$- \frac{4 x^{3}}{6 y^{- 3} - 1}$ $-(4x^3)/(6y^(-3)-1)$

Here,
$x^{4} - \frac{3}{y^{2}} - y = 8$ $x^4-3/(y^2)-y=8$

now lets differentiate the two sides for x

$\frac{d}{d x} (x^{4}) - \frac{d}{d x} (3 y^{- 2}) - \frac{d y}{d x} = 0$ $d/(dx)(x^4)-d/(dx)(3y^(-2))-(dy)/(dx)=0$

$or, 4 x^{3} + 6 y^{- 3} \frac{d y}{d x} - \frac{d y}{d x} = 0$ $or,4x^3+6y^(-3)(dy)/(dx)-(dy)/(dx)=0$

$or, (6 y^{- 3} - 1) \frac{d y}{d x} = - 4 x^{3}$ $or,(6y^(-3)-1)(dy)/(dx)=-4x^3$

$or, \frac{d y}{d x} = - \frac{4 x^{3}}{6 y^{- 3} - 1}$ $or,(dy)/(dx)=-(4x^3)/(6y^(-3)-1)$

How do you implicitly differentiate x^4-3/y^2-y=8 x4−3y2−y=8x^4-3/y^2-y=8 ?