What is the derivative of f(x)= 1/((x^3+7)^4) ?

The derivative of f(x)= 1/((x^3+7)^4) is -(12x^2)/((x^3+7)^5)

What is the first derivative of f(x)= 1/((x^3+7)^4) ?

The first derivative of f(x)= 1/((x^3+7)^4) is -(12x^2)/((x^3+7)^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) = \frac{1}{( x ^{3} + 7 ) ^{4}}

- \frac{12 x ^{2}}{( x ^{3} + 7 ) ^{5}}

Solution steps

\frac{d}{d x} (\frac{1}{( x ^{3} + 7 ) ^{4}})

Apply exponent rule:

\frac{1}{a} = a^{- 1}

= \frac{d}{d x} ((x^{3} + 7)^{- 4})

Apply the chain rule:

- \frac{4}{( x ^{3} + 7 ) ^{5}} \frac{d}{d x} (x^{3} + 7)

= - \frac{4}{( x ^{3} + 7 ) ^{5}} \frac{d}{d x} (x^{3} + 7)

\frac{d}{d x} (x^{3} + 7) = 3 x^{2}

= - \frac{4}{( x ^{3} + 7 ) ^{5}} \cdot 3 x^{2}

Simplify

- \frac{4}{( x ^{3} + 7 ) ^{5}} \cdot 3 x^{2} : - \frac{12 x ^{2}}{( x ^{3} + 7 ) ^{5}}

= - \frac{12 x ^{2}}{( x ^{3} + 7 ) ^{5}}

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

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

n = 1 \sum \infty \frac{n}{5 + n ^{2}}

2 x - x^{2}, - 2 x

\int \frac{1 + 16 x}{25 - x ^{2}} d x

What is the derivative of f(x)= 1/((x^3+7)^4) ?
The derivative of f(x)= 1/((x^3+7)^4) is -(12x^2)/((x^3+7)^5)
What is the first derivative of f(x)= 1/((x^3+7)^4) ?
The first derivative of f(x)= 1/((x^3+7)^4) is -(12x^2)/((x^3+7)^5)