What is the derivative of 4x^4-32x^3+89x^2-95x+29 ?

The derivative of 4x^4-32x^3+89x^2-95x+29 is 16x^3-96x^2+178x-95

What is the first derivative of 4x^4-32x^3+89x^2-95x+29 ?

The first derivative of 4x^4-32x^3+89x^2-95x+29 is 16x^3-96x^2+178x-95

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

\frac{d}{d x} (4 x^{4} - 32 x^{3} + 89 x^{2} - 95 x + 29)

16 x^{3} - 96 x^{2} + 178 x - 95

Solution steps

\frac{d}{d x} (4 x^{4} - 32 x^{3} + 89 x^{2} - 95 x + 29)

Apply the Sum/Difference Rule:

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

= \frac{d}{d x} (4 x^{4}) - \frac{d}{d x} (32 x^{3}) + \frac{d}{d x} (89 x^{2}) - \frac{d}{d x} (95 x) + \frac{d}{d x} (29)

\frac{d}{d x} (4 x^{4}) = 16 x^{3}

\frac{d}{d x} (32 x^{3}) = 96 x^{2}

\frac{d}{d x} (89 x^{2}) = 178 x

\frac{d}{d x} (95 x) = 95

\frac{d}{d x} (29) = 0

= 16 x^{3} - 96 x^{2} + 178 x - 95 + 0

Simplify

= 16 x^{3} - 96 x^{2} + 178 x - 95

x \to - 5 + lim (\frac{x + 4}{x + 5})

\int_{0}^{2} 4 x - 2 x^{2} d x

t + sin (t)

sin (2 t) cos (5 t)

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

What is the derivative of 4x^4-32x^3+89x^2-95x+29 ?
The derivative of 4x^4-32x^3+89x^2-95x+29 is 16x^3-96x^2+178x-95
What is the first derivative of 4x^4-32x^3+89x^2-95x+29 ?
The first derivative of 4x^4-32x^3+89x^2-95x+29 is 16x^3-96x^2+178x-95