What is the derivative of f(x)= 5/(x^3)-4/(x^2)+1/x ?

The derivative of f(x)= 5/(x^3)-4/(x^2)+1/x is -(15)/(x^4)+8/(x^3)-1/(x^2)

What is the first derivative of f(x)= 5/(x^3)-4/(x^2)+1/x ?

The first derivative of f(x)= 5/(x^3)-4/(x^2)+1/x is -(15)/(x^4)+8/(x^3)-1/(x^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{5}{x ^{3}} - \frac{4}{x ^{2}} + \frac{1}{x}

- \frac{15}{x ^{4}} + \frac{8}{x ^{3}} - \frac{1}{x ^{2}}

Solution steps

\frac{d}{d x} (\frac{5}{x ^{3}} - \frac{4}{x ^{2}} + \frac{1}{x})

Apply the Sum/Difference Rule:

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

= \frac{d}{d x} (\frac{5}{x ^{3}}) - \frac{d}{d x} (\frac{4}{x ^{2}}) + \frac{d}{d x} (\frac{1}{x})

\frac{d}{d x} (\frac{5}{x ^{3}}) = - \frac{15}{x ^{4}}

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

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

= - \frac{15}{x ^{4}} - (- \frac{8}{x ^{3}}) - \frac{1}{x ^{2}}

Simplify

= - \frac{15}{x ^{4}} + \frac{8}{x ^{3}} - \frac{1}{x ^{2}}

\frac{d}{d x} (2 tan (x) - 4 sec (x))

x \to \frac{3 π}{2} lim (tan (x))

\int x^{3} e^{4 x^{4}} d x

x \to 1 - lim (\frac{1}{x ^{2} + 1})

What is the derivative of f(x)= 5/(x^3)-4/(x^2)+1/x ?
The derivative of f(x)= 5/(x^3)-4/(x^2)+1/x is -(15)/(x^4)+8/(x^3)-1/(x^2)
What is the first derivative of f(x)= 5/(x^3)-4/(x^2)+1/x ?
The first derivative of f(x)= 5/(x^3)-4/(x^2)+1/x is -(15)/(x^4)+8/(x^3)-1/(x^2)