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

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

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

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

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derivative of

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

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

Solution steps

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

Take the constant out:

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

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

Apply exponent rule:

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

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

Apply the Power Rule:

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

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

Simplify

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

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

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

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

\int cos (3 t) sin (3 t) d t

\int \frac{12 x ^{2}}{4 x ^{3} - 1} d x

\int \frac{4 x + 7}{( 8 x ^{2} + 28 x ) ^{3}} d x

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