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

The derivative of 3/(2sqrt(3x+1)) is -9/(4(3x+1)^{3/2)}

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

The first derivative of 3/(2sqrt(3x+1)) is -9/(4(3x+1)^{3/2)}

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

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

- \frac{9}{4 ( 3 x + 1 ) ^{\frac{3}{2}}}

Solution steps

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

Take the constant out:

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

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

Apply exponent rule:

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

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

Simplify

\frac{3}{2} (- \frac{1}{2 ( 3 x + 1 ) ^{\frac{3}{2}}} \cdot 3) : - \frac{9}{4 ( 3 x + 1 ) ^{\frac{3}{2}}}

= - \frac{9}{4 ( 3 x + 1 ) ^{\frac{3}{2}}}

x \to 0 lim (4 cos (2 x))

y^{^{''}} - y^{^{'}} = 6 x e^{x}

x \to 0 lim (x^{3} - 3 x^{2} + 2 x - 3)

2 y^{^{''}} - 32 y = 0

What is the derivative of 3/(2sqrt(3x+1)) ?
The derivative of 3/(2sqrt(3x+1)) is -9/(4(3x+1)^{3/2)}
What is the first derivative of 3/(2sqrt(3x+1)) ?
The first derivative of 3/(2sqrt(3x+1)) is -9/(4(3x+1)^{3/2)}