What is the derivative of 1/(x^2y) ?

The derivative of 1/(x^2y) is -(2y+x\frac{dy)/(dx)}{x^3y^2}

What is the first derivative of 1/(x^2y) ?

The first derivative of 1/(x^2y) is -(2y+x\frac{dy)/(dx)}{x^3y^2}

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\frac{d}{d x} (\frac{1}{x ^{2} y})

- \frac{2 y + x \frac{d y}{d x}}{x ^{3} y ^{2}}

Solution steps

\frac{d}{d x} (\frac{1}{x ^{2} y})

Apply exponent rule:

\frac{1}{a} = a^{- 1}

= \frac{d}{d x} ((x^{2} y)^{- 1})

Apply the chain rule:

- \frac{1}{( x ^{2} y ) ^{2}} \frac{d}{d x} (x^{2} y)

= - \frac{1}{( x ^{2} y ) ^{2}} \frac{d}{d x} (x^{2} y)

\frac{d}{d x} (x^{2} y) = 2 x y + \frac{d y}{d x} x^{2}

= - \frac{1}{( x ^{2} y ) ^{2}} (2 x y + \frac{d y}{d x} x^{2})

Simplify

- \frac{1}{( x ^{2} y ) ^{2}} (2 x y + \frac{d y}{d x} x^{2}) : - \frac{2 y + x \frac{d y}{d x}}{x ^{3} y ^{2}}

= - \frac{2 y + x \frac{d y}{d x}}{x ^{3} y ^{2}}

\int_{0}^{1} x^{2} (x + 1)^{7} d x

\frac{\partial}{\partial y} (2 x y^{3} - 4)

y^{^{''}} - 18 y^{^{'}} + 81 y = 0

\int 16 tan^{3} (x) d x

g (w) = 10 w

What is the derivative of 1/(x^2y) ?
The derivative of 1/(x^2y) is -(2y+x\frac{dy)/(dx)}{x^3y^2}
What is the first derivative of 1/(x^2y) ?
The first derivative of 1/(x^2y) is -(2y+x\frac{dy)/(dx)}{x^3y^2}