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

The derivative of f(x)= 2/(3x+1) is -6/((3x+1)^2)

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

The first derivative of f(x)= 2/(3x+1) is -6/((3x+1)^2)

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derivative of

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

- \frac{6}{( 3 x + 1 ) ^{2}}

Solution steps

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

Take the constant out:

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

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

Apply exponent rule:

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

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

Apply the chain rule:

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

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

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

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

Simplify

2 (- \frac{1}{( 3 x + 1 ) ^{2}} \cdot 3) : - \frac{6}{( 3 x + 1 ) ^{2}}

= - \frac{6}{( 3 x + 1 ) ^{2}}

5 sin (t)

\int_{0}^{1} \frac{1}{e ^{x} + 1} d x

a y^{^{'}} = b x - b y

\frac{d y}{d x} + (5 x) y = 3 x + 5

\int e^{t} + e^{- t} d t

What is the derivative of f(x)= 2/(3x+1) ?
The derivative of f(x)= 2/(3x+1) is -6/((3x+1)^2)
What is the first derivative of f(x)= 2/(3x+1) ?
The first derivative of f(x)= 2/(3x+1) is -6/((3x+1)^2)