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

The derivative of-(16/((1+4x)^2)) is (128)/((1+4x)^3)

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

The first derivative of-(16/((1+4x)^2)) is (128)/((1+4x)^3)

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

\frac{128}{( 1 + 4 x ) ^{3}}

Solution steps

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

Take the constant out:

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

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

Apply exponent rule:

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

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

Apply the chain rule:

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

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

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

= - 16 (- \frac{2}{( 1 + 4 x ) ^{3}} \cdot 4)

Simplify

- 16 (- \frac{2}{( 1 + 4 x ) ^{3}} \cdot 4) : \frac{128}{( 1 + 4 x ) ^{3}}

= \frac{128}{( 1 + 4 x ) ^{3}}

\int_{1}^{2} 21 t^{2} d t

\frac{d}{d x} (x - 8)

4^{\frac{x}{l n ( 2 )}}, at x = ln (16)

\int t (t^{2} + 1) d t

(3 x y + y^{2}) d x + (x^{2} + x y) d y = 0

What is the derivative of-(16/((1+4x)^2)) ?
The derivative of-(16/((1+4x)^2)) is (128)/((1+4x)^3)
What is the first derivative of-(16/((1+4x)^2)) ?
The first derivative of-(16/((1+4x)^2)) is (128)/((1+4x)^3)