What is the derivative of f(x)= 6/(3x+1),\at a=-1 ?

The derivative of f(x)= 6/(3x+1),\at a=-1 is -(18)/((3x+1)^2)

What is the first derivative of f(x)= 6/(3x+1),\at a=-1 ?

The first derivative of f(x)= 6/(3x+1),\at a=-1 is -(18)/((3x+1)^2)

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derivative of

f (x) = \frac{6}{3 x + 1}, at a = - 1

- \frac{18}{( 3 x + 1 ) ^{2}}

Solution steps

\frac{d}{d x} (\frac{6}{3 x + 1})

Take the constant out:

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

= 6 \frac{d}{d x} (\frac{1}{3 x + 1})

Apply exponent rule:

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

= 6 \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

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

Simplify

6 (- \frac{1}{( 3 x + 1 ) ^{2}} \cdot 3) : - \frac{18}{( 3 x + 1 ) ^{2}}

= - \frac{18}{( 3 x + 1 ) ^{2}}

\int \frac{x}{1 + x} d x

\frac{d}{d x} (- \frac{1}{4 x ^{4}})

\int cos (x) sin (x) d x

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

f (x) = cos (x^{2}) \cdot 2 x

What is the derivative of f(x)= 6/(3x+1),\at a=-1 ?
The derivative of f(x)= 6/(3x+1),\at a=-1 is -(18)/((3x+1)^2)
What is the first derivative of f(x)= 6/(3x+1),\at a=-1 ?
The first derivative of f(x)= 6/(3x+1),\at a=-1 is -(18)/((3x+1)^2)