What is the derivative of y=(x^2+4x+3)/(sqrt(x)) ?

The derivative of y=(x^2+4x+3)/(sqrt(x)) is (3x^2+4x-3)/(2xsqrt(x))

What is the first derivative of y=(x^2+4x+3)/(sqrt(x)) ?

The first derivative of y=(x^2+4x+3)/(sqrt(x)) is (3x^2+4x-3)/(2xsqrt(x))

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

derivative of

y = \frac{x ^{2} + 4 x + 3}{x}

\frac{3 x ^{2} + 4 x - 3}{2 x x}

Solution steps

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

Apply the Quotient Rule:

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

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

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

\frac{d}{d x} (x) = \frac{1}{2 x}

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

Simplify

\frac{( 2 x + 4 ) x - \frac{1}{2 x} ( x ^{2} + 4 x + 3 )}{( x ) ^{2}} : \frac{3 x ^{2} + 4 x - 3}{2 x x}

= \frac{3 x ^{2} + 4 x - 3}{2 x x}

f (x) = \frac{1 - sec ( x )}{tan ( x )}

x = - 2

y = 2 x - 5

f (x) = x e^{- x^{2}}

y = tan (x)

What is the derivative of y=(x^2+4x+3)/(sqrt(x)) ?
The derivative of y=(x^2+4x+3)/(sqrt(x)) is (3x^2+4x-3)/(2xsqrt(x))
What is the first derivative of y=(x^2+4x+3)/(sqrt(x)) ?
The first derivative of y=(x^2+4x+3)/(sqrt(x)) is (3x^2+4x-3)/(2xsqrt(x))