What is (\partial)/(\partial x)(y-4x^2y^{-2}) ?

The answer to (\partial)/(\partial x)(y-4x^2y^{-2}) is -(8x)/(y^2)

(\partial)/(\partial x)(y-4x^2y^{-2})

\frac{\partial}{\partial x} (y - 4 x^{2} y^{- 2})

- \frac{8 x}{y ^{2}}

Solution steps

\frac{\partial}{\partial x} (y - 4 x^{2} y^{- 2})

Treat

y

as a constant

Apply the Sum/Difference Rule:

(f \pm g)^{'} = f^{'} \pm g^{'}

= \frac{\partial}{\partial x} (y) - \frac{\partial}{\partial x} (4 x^{2} y^{- 2})

\frac{\partial}{\partial x} (y) = 0

\frac{\partial}{\partial x} (4 x^{2} y^{- 2}) = \frac{8 x}{y ^{2}}

= 0 - \frac{8 x}{y ^{2}}

Simplify

= - \frac{8 x}{y ^{2}}

\frac{\partial ^{2}}{\partial x ^{2}} (ln (3 + x^{2} y^{2}))

\int_{0}^{l n (x)} sin (e^{t}) d t

n = 3 \sum \infty \frac{8}{n ^{2} - 1}

d er i v a t i v e 4 x^{3} + 6 x^{2}

\frac{\partial}{\partial x} (6 x + 3 y - 7)

What is (\partial)/(\partial x)(y-4x^2y^{-2}) ?
The answer to (\partial)/(\partial x)(y-4x^2y^{-2}) is -(8x)/(y^2)