What is the derivative of x/8+(sqrt((6-x^2+4))/3) ?

The derivative of x/8+(sqrt((6-x^2+4))/3) is 1/8+(x-6)/(3sqrt(x^2-12x+40))

What is the first derivative of x/8+(sqrt((6-x^2+4))/3) ?

The first derivative of x/8+(sqrt((6-x^2+4))/3) is 1/8+(x-6)/(3sqrt(x^2-12x+40))

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

\frac{1}{8} + \frac{x - 6}{3 x ^{2} - 12 x + 40}

Solution steps

\frac{d}{d x} \frac{x}{8} + \frac{( 6 - x ) ^{2} + 4}{3}

Simplify

\frac{x}{8} + \frac{( 6 - x ) ^{2} + 4}{3} : \frac{x}{8} + \frac{x ^{2} - 12 x + 40}{3}

= \frac{d}{d x} (\frac{x}{8} + \frac{x ^{2} - 12 x + 40}{3})

Apply the Sum/Difference Rule:

(f \pm g)^{'} = f^{'} \pm g^{'}

= \frac{d}{d x} (\frac{x}{8}) + \frac{d}{d x} (\frac{x ^{2} - 12 x + 40}{3})

\frac{d}{d x} (\frac{x}{8}) = \frac{1}{8}

\frac{d}{d x} (\frac{x ^{2} - 12 x + 40}{3}) = \frac{x - 6}{3 x ^{2} - 12 x + 40}

= \frac{1}{8} + \frac{x - 6}{3 x ^{2} - 12 x + 40}

x \to 0 lim (\frac{sin ( \frac{x}{3} )}{x})

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

\frac{d}{d x} (a + b x)

\int \frac{4 x + 3}{x ^{2} + 9} d x

\frac{d ^{2}}{d x ^{2}} (\frac{7}{8 x ^{5}})

What is the derivative of x/8+(sqrt((6-x^2+4))/3) ?
The derivative of x/8+(sqrt((6-x^2+4))/3) is 1/8+(x-6)/(3sqrt(x^2-12x+40))
What is the first derivative of x/8+(sqrt((6-x^2+4))/3) ?
The first derivative of x/8+(sqrt((6-x^2+4))/3) is 1/8+(x-6)/(3sqrt(x^2-12x+40))