What is the derivative of (3x^4-x+1(-x^5+4)) ?

The derivative of (3x^4-x+1(-x^5+4)) is -27x^8+6x^5-5x^4+48x^3-4

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

The first derivative of (3x^4-x+1(-x^5+4)) is -27x^8+6x^5-5x^4+48x^3-4

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

\frac{d}{d x} ((3 x^{4} - x + 1) (- x^{5} + 4))

- 27 x^{8} + 6 x^{5} - 5 x^{4} + 48 x^{3} - 4

Solution steps

\frac{d}{d x} ((3 x^{4} - x + 1) (- x^{5} + 4))

Apply the Product Rule:

(f \cdot g)^{'} = f^{'} \cdot g + f \cdot g^{'}

f = 3 x^{4} - x + 1, g = - x^{5} + 4

= \frac{d}{d x} (3 x^{4} - x + 1) (- x^{5} + 4) + \frac{d}{d x} (- x^{5} + 4) (3 x^{4} - x + 1)

\frac{d}{d x} (3 x^{4} - x + 1) = 12 x^{3} - 1

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

= (12 x^{3} - 1) (- x^{5} + 4) + (- 5 x^{4}) (3 x^{4} - x + 1)

Simplify

(12 x^{3} - 1) (- x^{5} + 4) + (- 5 x^{4}) (3 x^{4} - x + 1) : - 27 x^{8} + 6 x^{5} - 5 x^{4} + 48 x^{3} - 4

= - 27 x^{8} + 6 x^{5} - 5 x^{4} + 48 x^{3} - 4

y^{^{'}} + 2 y = 1, y (0) = \frac{5}{2}

\int x (2 x + 3)^{2} d x

\frac{d y}{d x} = \frac{y cos ( x )}{1 - 2 y}

\frac{d}{d x} (12)

\frac{d}{d x} (x ln (cos (x)))

What is the derivative of (3x^4-x+1(-x^5+4)) ?
The derivative of (3x^4-x+1(-x^5+4)) is -27x^8+6x^5-5x^4+48x^3-4
What is the first derivative of (3x^4-x+1(-x^5+4)) ?
The first derivative of (3x^4-x+1(-x^5+4)) is -27x^8+6x^5-5x^4+48x^3-4