What is the derivative of f(x)=(x^4-5/(x^3))^{-4} ?

The derivative of f(x)=(x^4-5/(x^3))^{-4} is -(4x^{11}(4x^7+15))/((x^7-5)^5)

What is the first derivative of f(x)=(x^4-5/(x^3))^{-4} ?

The first derivative of f(x)=(x^4-5/(x^3))^{-4} is -(4x^{11}(4x^7+15))/((x^7-5)^5)

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

derivative of

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

- \frac{4 x ^{11} ( 4 x ^{7} + 15 )}{( x ^{7} - 5 ) ^{5}}

Solution steps

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

Apply the chain rule:

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

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

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

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

Simplify

- \frac{4}{( x ^{4} - \frac{5}{x ^{3}} ) ^{5}} (4 x^{3} + \frac{15}{x ^{4}}) : - \frac{4 x ^{11} ( 4 x ^{7} + 15 )}{( x ^{7} - 5 ) ^{5}}

= - \frac{4 x ^{11} ( 4 x ^{7} + 15 )}{( x ^{7} - 5 ) ^{5}}

\int_{0}^{3} \frac{x}{x + 1} d x

n = 1 \sum \infty \frac{2 n}{3 n - 2}

\frac{d y}{d x} = (y - 1) x

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

(arctan (x))^{^{'}}

What is the derivative of f(x)=(x^4-5/(x^3))^{-4} ?
The derivative of f(x)=(x^4-5/(x^3))^{-4} is -(4x^{11}(4x^7+15))/((x^7-5)^5)
What is the first derivative of f(x)=(x^4-5/(x^3))^{-4} ?
The first derivative of f(x)=(x^4-5/(x^3))^{-4} is -(4x^{11}(4x^7+15))/((x^7-5)^5)