Find the Derivative minus d First, divide dx four divide(three x to the power of two)

$\frac{4}{3 x^{2}}$

Since $\frac{4}{3}$ is constant with respect to $x$ , the derivative of $\frac{4}{3 x^{2}}$ with respect to $x$ is $\frac{4}{3} \frac{d}{d x} [\frac{1}{x^{2}}]$ .

$\frac{4}{3} \frac{d}{d x} [\frac{1}{x^{2}}]$

Apply basic rules of exponents.

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Should be rewritten $\frac{1}{x^{2}}$ as ${(x^{2})}^{- 1}$ .

$\frac{4}{3} \frac{d}{d x} [{(x^{2})}^{- 1}]$

First, multiply the exponents in ${(x^{2})}^{- 1}$ .

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Apply the power rule when multiplying the exponents, ${(a^{m})}^{n} = a^{m n}$ .

$\frac{4}{3} \frac{d}{d x} [x^{2 \cdot - 1}]$

First, multiply $2$ by $- 1$ .

$\frac{4}{3} \frac{d}{d x} [x^{- 2}]$

Differentiate using the Power Rule which states that $\frac{d}{d x} [x^{n}]$ is $n x^{n - 1}$ where $n = - 2$ .

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

Add fractions.

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Add $- 2$ and $\frac{4}{3}$ .

$\frac{- 2 \cdot 4}{3} x^{- 3}$

Multiply $- 2$ by $4$ .

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

Combine $\frac{- 8}{3}$ and $x^{- 3}$ .

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

Clarify the expression.

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Move $x^{- 3}$ to the denominator using the negative exponent rule $b^{- n} = \frac{1}{b^{n}}$ .

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

Put the negative in front of the fraction.

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

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