What is the derivative of 1/((x^3-2x^2+7^4)) ?

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

What is the first derivative of 1/((x^3-2x^2+7^4)) ?

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

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{1}{( x ^{3} - 2 x ^{2} + 7 ) ^{4}})

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

Solution steps

\frac{d}{d x} (\frac{1}{( x ^{3} - 2 x ^{2} + 7 ) ^{4}})

Apply exponent rule:

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

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

Apply the chain rule:

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

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

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

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

Multiply fractions:

a \cdot \frac{b}{c} = \frac{a \cdot b}{c}

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

n = 1 \sum \infty 4 n - 1

\int \frac{1}{1 + 3 x ^{2}} d x

\frac{d}{d x} (\frac{3 x}{s q u ( x , q ) a r ( x , q ) e x ^{2} + 1})

x \to 1 lim (\frac{1}{x ^{3}})

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