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

The derivative of y= 2/(3x^4) is -8/(3x^5)

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

The first derivative of y= 2/(3x^4) is -8/(3x^5)

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derivative of

y = \frac{2}{3 x ^{4}}

- \frac{8}{3 x ^{5}}

Solution steps

\frac{d}{d x} (\frac{2}{3 x ^{4}})

Take the constant out:

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

= \frac{2}{3} \frac{d}{d x} (\frac{1}{x ^{4}})

Apply exponent rule:

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

= \frac{2}{3} \frac{d}{d x} (x^{- 4})

Apply the Power Rule:

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

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

Simplify

\frac{2}{3} (- 4 x^{- 4 - 1}) : - \frac{8}{3 x ^{5}}

= - \frac{8}{3 x ^{5}}

e^{2 x} (2 x + 3)^{7}

\frac{\partial}{\partial t} (x)

x d y = (2 x e^{x} - y + 6 x^{2}) d x

\int 2 e^{x} d x

\int (3 x + 1)^{2} d x

What is the derivative of y= 2/(3x^4) ?
The derivative of y= 2/(3x^4) is -8/(3x^5)
What is the first derivative of y= 2/(3x^4) ?
The first derivative of y= 2/(3x^4) is -8/(3x^5)