What is the derivative of y= 2/(3x^2) ?

The derivative of y= 2/(3x^2) is -4/(3x^3)

What is the first derivative of y= 2/(3x^2) ?

The first derivative of y= 2/(3x^2) is -4/(3x^3)

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derivative of

y = \frac{2}{3 x ^{2}}

- \frac{4}{3 x ^{3}}

Solution steps

\frac{d}{d x} (\frac{2}{3 x ^{2}})

Take the constant out:

(a \cdot f)^{'} = a \cdot f^{'}

= \frac{2}{3} \frac{d}{d x} (\frac{1}{x ^{2}})

Apply exponent rule:

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

= \frac{2}{3} \frac{d}{d x} (x^{- 2})

Apply the Power Rule:

\frac{d}{d x} (x^{a}) = a \cdot x^{a - 1}

= \frac{2}{3} (- 2 x^{- 2 - 1})

Simplify

\frac{2}{3} (- 2 x^{- 2 - 1}) : - \frac{4}{3 x ^{3}}

= - \frac{4}{3 x ^{3}}

x \to 0 lim (\frac{4 x ^{2}}{x})

y^{^{'}} = 9 y^{2} + x y^{2}, y (0) = 1

\frac{\partial}{\partial x} ((41 - x^{2} - y^{2})^{\frac{1}{2}})

\int_{0}^{1} x e^{- 4 x} d x

\int \frac{1}{x 25 x ^{2} + 16} d x

What is the derivative of y= 2/(3x^2) ?
The derivative of y= 2/(3x^2) is -4/(3x^3)
What is the first derivative of y= 2/(3x^2) ?
The first derivative of y= 2/(3x^2) is -4/(3x^3)