What is the derivative of f(x)=(64)/((1+4x)^3) ?

The derivative of f(x)=(64)/((1+4x)^3) is -(768)/((1+4x)^4)

What is the first derivative of f(x)=(64)/((1+4x)^3) ?

The first derivative of f(x)=(64)/((1+4x)^3) is -(768)/((1+4x)^4)

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derivative of

f (x) = \frac{64}{( 1 + 4 x ) ^{3}}

- \frac{768}{( 1 + 4 x ) ^{4}}

Solution steps

\frac{d}{d x} (\frac{64}{( 1 + 4 x ) ^{3}})

Take the constant out:

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

= 64 \frac{d}{d x} (\frac{1}{( 1 + 4 x ) ^{3}})

Apply exponent rule:

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

= 64 \frac{d}{d x} ((1 + 4 x)^{- 3})

Apply the chain rule:

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

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

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

= 64 (- \frac{3}{( 1 + 4 x ) ^{4}} \cdot 4)

Simplify

64 (- \frac{3}{( 1 + 4 x ) ^{4}} \cdot 4) : - \frac{768}{( 1 + 4 x ) ^{4}}

= - \frac{768}{( 1 + 4 x ) ^{4}}

y^{^{'}} = cos^{2} (x) cos^{2} (2 y)

\int \frac{x}{x ^{4} + 49} d x

d er i v a t i v e (x + i y)^{2}

y^{^{'}} + (\frac{2}{x}) y = 0

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

What is the derivative of f(x)=(64)/((1+4x)^3) ?
The derivative of f(x)=(64)/((1+4x)^3) is -(768)/((1+4x)^4)
What is the first derivative of f(x)=(64)/((1+4x)^3) ?
The first derivative of f(x)=(64)/((1+4x)^3) is -(768)/((1+4x)^4)