What is the derivative of f(x)=(6x^5+4x)/(x^4+3) ?

The derivative of f(x)=(6x^5+4x)/(x^4+3) is (6x^8+78x^4+12)/((x^4+3)^2)

What is the first derivative of f(x)=(6x^5+4x)/(x^4+3) ?

The first derivative of f(x)=(6x^5+4x)/(x^4+3) is (6x^8+78x^4+12)/((x^4+3)^2)

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{6 x ^{5} + 4 x}{x ^{4} + 3}

\frac{6 x ^{8} + 78 x ^{4} + 12}{( x ^{4} + 3 ) ^{2}}

Solution steps

\frac{d}{d x} (\frac{6 x ^{5} + 4 x}{x ^{4} + 3})

Apply the Quotient Rule:

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

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

\frac{d}{d x} (6 x^{5} + 4 x) = 30 x^{4} + 4

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

= \frac{( 30 x ^{4} + 4 ) ( x ^{4} + 3 ) - 4 x ^{3} ( 6 x ^{5} + 4 x )}{( x ^{4} + 3 ) ^{2}}

Expand

(30 x^{4} + 4) (x^{4} + 3) - 4 x^{3} (6 x^{5} + 4 x) : 6 x^{8} + 78 x^{4} + 12

= \frac{6 x ^{8} + 78 x ^{4} + 12}{( x ^{4} + 3 ) ^{2}}

\int (2 x - 3) d x

\int cot^{5} (4 x) d x

\frac{1}{16 + x ^{2}}

n = 1 \sum \infty (\frac{4}{10})^{n}

\int \frac{2 x + 2}{x ^{2} - 4 x + 9} d x

What is the derivative of f(x)=(6x^5+4x)/(x^4+3) ?
The derivative of f(x)=(6x^5+4x)/(x^4+3) is (6x^8+78x^4+12)/((x^4+3)^2)
What is the first derivative of f(x)=(6x^5+4x)/(x^4+3) ?
The first derivative of f(x)=(6x^5+4x)/(x^4+3) is (6x^8+78x^4+12)/((x^4+3)^2)