What is the derivative of f(x)=(1+16x^2)/((4x)^3) ?

The derivative of f(x)=(1+16x^2)/((4x)^3) is (-16x^2-3)/(64x^4)

What is the first derivative of f(x)=(1+16x^2)/((4x)^3) ?

The first derivative of f(x)=(1+16x^2)/((4x)^3) is (-16x^2-3)/(64x^4)

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

derivative of

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

\frac{- 16 x ^{2} - 3}{64 x ^{4}}

Solution steps

\frac{d}{d x} (\frac{1 + 16 x ^{2}}{( 4 x ) ^{3}})

Apply the Quotient Rule:

(\frac{f}{g})^{'} = \frac{f ^{'} \cdot g - g ^{'} \cdot f}{g ^{2}}

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

\frac{d}{d x} (1 + 16 x^{2}) = 32 x

\frac{d}{d x} ((4 x)^{3}) = 192 x^{2}

= \frac{32 x ( 4 x ) ^{3} - 192 x ^{2} ( 1 + 16 x ^{2} )}{( ( 4 x ) ^{3} ) ^{2}}

Simplify

\frac{32 x ( 4 x ) ^{3} - 192 x ^{2} ( 1 + 16 x ^{2} )}{( ( 4 x ) ^{3} ) ^{2}} : \frac{- 16 x ^{2} - 3}{64 x ^{4}}

= \frac{- 16 x ^{2} - 3}{64 x ^{4}}

y^{^{''}} - y^{^{'}} = 0, y (0) = 10

\int \frac{4}{3 - x} d x

\frac{\partial}{\partial x} (x^{2} - x y - y^{2})

\frac{d y}{d x} = x y + 3 x + 5 y + 15

\frac{d}{d x} (\frac{5 ^{x}}{1 + 5 ^{2 x}})

What is the derivative of f(x)=(1+16x^2)/((4x)^3) ?
The derivative of f(x)=(1+16x^2)/((4x)^3) is (-16x^2-3)/(64x^4)
What is the first derivative of f(x)=(1+16x^2)/((4x)^3) ?
The first derivative of f(x)=(1+16x^2)/((4x)^3) is (-16x^2-3)/(64x^4)