What is the derivative of (2x^3+5/(4x^2+7)) ?

The derivative of (2x^3+5/(4x^2+7)) is (8x^4+42x^2-40x)/((4x^2+7)^2)

What is the first derivative of (2x^3+5/(4x^2+7)) ?

The first derivative of (2x^3+5/(4x^2+7)) is (8x^4+42x^2-40x)/((4x^2+7)^2)

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

\frac{d}{d x} (\frac{2 x ^{3} + 5}{4 x ^{2} + 7})

\frac{8 x ^{4} + 42 x ^{2} - 40 x}{( 4 x ^{2} + 7 ) ^{2}}

Solution steps

\frac{d}{d x} (\frac{2 x ^{3} + 5}{4 x ^{2} + 7})

Apply the Quotient Rule:

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

= \frac{\frac{d}{d x} ( 2 x ^{3} + 5 ) ( 4 x ^{2} + 7 ) - \frac{d}{d x} ( 4 x ^{2} + 7 ) ( 2 x ^{3} + 5 )}{( 4 x ^{2} + 7 ) ^{2}}

\frac{d}{d x} (2 x^{3} + 5) = 6 x^{2}

\frac{d}{d x} (4 x^{2} + 7) = 8 x

= \frac{6 x ^{2} ( 4 x ^{2} + 7 ) - 8 x ( 2 x ^{3} + 5 )}{( 4 x ^{2} + 7 ) ^{2}}

Expand

6 x^{2} (4 x^{2} + 7) - 8 x (2 x^{3} + 5) : 8 x^{4} + 42 x^{2} - 40 x

= \frac{8 x ^{4} + 42 x ^{2} - 40 x}{( 4 x ^{2} + 7 ) ^{2}}

\frac{d}{d x} ((f (x) (x^{3}))^{2})

\frac{d}{d x} (3 cot (2 x) sec (2 x))

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

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

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

What is the derivative of (2x^3+5/(4x^2+7)) ?
The derivative of (2x^3+5/(4x^2+7)) is (8x^4+42x^2-40x)/((4x^2+7)^2)
What is the first derivative of (2x^3+5/(4x^2+7)) ?
The first derivative of (2x^3+5/(4x^2+7)) is (8x^4+42x^2-40x)/((4x^2+7)^2)