What is the derivative of 4/(4x+1) ?

The derivative of 4/(4x+1) is -(16)/((4x+1)^2)

What is the first derivative of 4/(4x+1) ?

The first derivative of 4/(4x+1) is -(16)/((4x+1)^2)

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\frac{d}{d x} (\frac{4}{4 x + 1})

- \frac{16}{( 4 x + 1 ) ^{2}}

Solution steps

\frac{d}{d x} (\frac{4}{4 x + 1})

Take the constant out:

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

= 4 \frac{d}{d x} (\frac{1}{4 x + 1})

Apply exponent rule:

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

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

Apply the chain rule:

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

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

\frac{d}{d x} (4 x + 1) = 4

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

Simplify

4 (- \frac{1}{( 4 x + 1 ) ^{2}} \cdot 4) : - \frac{16}{( 4 x + 1 ) ^{2}}

= - \frac{16}{( 4 x + 1 ) ^{2}}

\int \frac{4 + 8 x}{1 + x ^{2}} d x

\int 9 x^{2} + 4 x - 3 d x

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

x \to 4 lim (2 x^{2} + 7)

\frac{d h}{d t} = \frac{( 7 - e ^{- \frac{t}{5}} )}{6} - \frac{2}{9} h

What is the derivative of 4/(4x+1) ?
The derivative of 4/(4x+1) is -(16)/((4x+1)^2)
What is the first derivative of 4/(4x+1) ?
The first derivative of 4/(4x+1) is -(16)/((4x+1)^2)