What is the derivative of 1/(16x^2) ?

The derivative of 1/(16x^2) is -1/(8x^3)

What is the first derivative of 1/(16x^2) ?

The first derivative of 1/(16x^2) is -1/(8x^3)

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\frac{d}{d x} (\frac{1}{16 x ^{2}})

- \frac{1}{8 x ^{3}}

Solution steps

\frac{d}{d x} (\frac{1}{16 x ^{2}})

Take the constant out:

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

= \frac{1}{16} \frac{d}{d x} (\frac{1}{x ^{2}})

Apply exponent rule:

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

= \frac{1}{16} \frac{d}{d x} (x^{- 2})

Apply the Power Rule:

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

= \frac{1}{16} (- 2 x^{- 2 - 1})

Simplify

\frac{1}{16} (- 2 x^{- 2 - 1}) : - \frac{1}{8 x ^{3}}

= - \frac{1}{8 x ^{3}}

\frac{d}{d x} (3 + 6 x)

\int_{0}^{9} x - \frac{1}{3} x d x

\frac{d y}{d t} = - 2 y

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

What is the derivative of 1/(16x^2) ?
The derivative of 1/(16x^2) is -1/(8x^3)
What is the first derivative of 1/(16x^2) ?
The first derivative of 1/(16x^2) is -1/(8x^3)