What is the derivative of y= 1/(3x) ?

The derivative of y= 1/(3x) is -1/(3x^2)

What is the first derivative of y= 1/(3x) ?

The first derivative of y= 1/(3x) is -1/(3x^2)

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derivative of

y = \frac{1}{3 x}

- \frac{1}{3 x ^{2}}

Solution steps

\frac{d}{d x} (\frac{1}{3 x})

Take the constant out:

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

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

Apply exponent rule:

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

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

Apply the Power Rule:

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

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

Simplify

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

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

\frac{d}{d x} (sin (x^{3} - x))

2 y = x, x + 1 = (y - 1)^{2}

\frac{d}{d x} (7 - 8 x)

\frac{d y}{d x} = (x + y + 5)^{2}

y^{^{''}} + 2 y = - 4 x

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