What is the derivative of ((sqrt(x^2-3x-4))/x) ?

The derivative of ((sqrt(x^2-3x-4))/x) is (3x+8)/(2x^2sqrt(x^2-3x-4))

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

The first derivative of ((sqrt(x^2-3x-4))/x) is (3x+8)/(2x^2sqrt(x^2-3x-4))

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{( x ^{2} - 3 x - 4 )}{x})

\frac{3 x + 8}{2 x ^{2} x ^{2} - 3 x - 4}

Solution steps

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

Apply the Quotient Rule:

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

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

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

\frac{d x}{d x} = 1

= \frac{\frac{2 x - 3}{2 x ^{2} - 3 x - 4} x - 1 \cdot x ^{2} - 3 x - 4}{x ^{2}}

Simplify

\frac{\frac{2 x - 3}{2 x ^{2} - 3 x - 4} x - 1 \cdot x ^{2} - 3 x - 4}{x ^{2}} : \frac{3 x + 8}{2 x ^{2} x ^{2} - 3 x - 4}

= \frac{3 x + 8}{2 x ^{2} x ^{2} - 3 x - 4}

x \to 0 lim (e^{a x})

\frac{t}{2 t - 9}

(6 x^{2} + 5)^{3}

\frac{d}{d x} (x + 2 cos (x))

\int sec (3 θ) tan (3 θ) d θ

What is the derivative of ((sqrt(x^2-3x-4))/x) ?
The derivative of ((sqrt(x^2-3x-4))/x) is (3x+8)/(2x^2sqrt(x^2-3x-4))
What is the first derivative of ((sqrt(x^2-3x-4))/x) ?
The first derivative of ((sqrt(x^2-3x-4))/x) is (3x+8)/(2x^2sqrt(x^2-3x-4))