What is the derivative of (x^3-3x^2+3x+1^'-(x-1)^') ?

The derivative of (x^3-3x^2+3x+1^'-(x-1)^') is '(3x^2-6x+3)(x^3-3x^2+3x+1)^{'-1}-'(x-1)^{'-1}

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

The first derivative of (x^3-3x^2+3x+1^'-(x-1)^') is '(3x^2-6x+3)(x^3-3x^2+3x+1)^{'-1}-'(x-1)^{'-1}

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

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

' (3 x^{2} - 6 x + 3) (x^{3} - 3 x^{2} + 3 x + 1)^{' - 1} -' (x - 1)^{' - 1}

Solution steps

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

Apply the Sum/Difference Rule:

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

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

\frac{d}{d x} ((x^{3} - 3 x^{2} + 3 x + 1)^{^{'}}) =^{'} (3 x^{2} - 6 x + 3) (x^{3} - 3 x^{2} + 3 x + 1)^{^{'} - 1}

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

=^{^{'}} (3 x^{2} - 6 x + 3) (x^{3} - 3 x^{2} + 3 x + 1)^{^{^{'}} - 1} -^{^{'}} (x - 1)^{^{^{'}} - 1}

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

6 (x^{2} + 3)^{- 1}

y = x^{2} \cdot (x + 1)^{3}

- 9 sin^{3} (x)

\frac{d}{dd} (ln (\frac{ln ( 8 l )}{d}) - 1)

What is the derivative of (x^3-3x^2+3x+1^'-(x-1)^') ?
The derivative of (x^3-3x^2+3x+1^'-(x-1)^') is '(3x^2-6x+3)(x^3-3x^2+3x+1)^{'-1}-'(x-1)^{'-1}
What is the first derivative of (x^3-3x^2+3x+1^'-(x-1)^') ?
The first derivative of (x^3-3x^2+3x+1^'-(x-1)^') is '(3x^2-6x+3)(x^3-3x^2+3x+1)^{'-1}-'(x-1)^{'-1}