What is the derivative of 1/(2sqrt(3x-1)) ?

The derivative of 1/(2sqrt(3x-1)) is -3/(4(3x-1)^{3/2)}

What is the first derivative of 1/(2sqrt(3x-1)) ?

The first derivative of 1/(2sqrt(3x-1)) is -3/(4(3x-1)^{3/2)}

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\frac{d}{d x} (\frac{1}{2 3 x - 1})

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

Solution steps

\frac{d}{d x} (\frac{1}{2 3 x - 1})

Take the constant out:

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

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

Apply exponent rule:

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

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

Apply the chain rule:

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

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

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

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

Simplify

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

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

f (x) = 8 (x - \frac{1}{x})^{4}, at x = 2

(sin (x^{'}))^{^{'}}

x - 2 + \frac{( x + 1 ) ^{2}}{2}

\int_{0}^{1} π (y - y^{2}) d y

What is the derivative of 1/(2sqrt(3x-1)) ?
The derivative of 1/(2sqrt(3x-1)) is -3/(4(3x-1)^{3/2)}
What is the first derivative of 1/(2sqrt(3x-1)) ?
The first derivative of 1/(2sqrt(3x-1)) is -3/(4(3x-1)^{3/2)}