What is the derivative of (18/((1+3x)^3)) ?

The derivative of (18/((1+3x)^3)) is -(162)/((1+3x)^4)

What is the first derivative of (18/((1+3x)^3)) ?

The first derivative of (18/((1+3x)^3)) is -(162)/((1+3x)^4)

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

- \frac{162}{( 1 + 3 x ) ^{4}}

Solution steps

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

Take the constant out:

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

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

Apply exponent rule:

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

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

Apply the chain rule:

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

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

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

= 18 (- \frac{3}{( 1 + 3 x ) ^{4}} \cdot 3)

Simplify

18 (- \frac{3}{( 1 + 3 x ) ^{4}} \cdot 3) : - \frac{162}{( 1 + 3 x ) ^{4}}

= - \frac{162}{( 1 + 3 x ) ^{4}}

\frac{25 s}{( s ^{2} + 9 ) ( s + 4 )}

1 + sin^{2} (x)

y^{^{'}} - \frac{2}{t} y = t^{2}

\frac{d}{d x} (- \frac{8 x}{( 4 x ^{2} + 1 ) ^{2}})

\frac{d}{d x} (arcsin (coth (x)))

What is the derivative of (18/((1+3x)^3)) ?
The derivative of (18/((1+3x)^3)) is -(162)/((1+3x)^4)
What is the first derivative of (18/((1+3x)^3)) ?
The first derivative of (18/((1+3x)^3)) is -(162)/((1+3x)^4)