What is the derivative of f(x)=(2x-8)^4(x^2+x+1)^5 ?

The derivative of f(x)=(2x-8)^4(x^2+x+1)^5 is 8(2x-8)^3(x^2+x+1)^5+5(x^2+x+1)^4(2x+1)(2x-8)^4

What is the first derivative of f(x)=(2x-8)^4(x^2+x+1)^5 ?

The first derivative of f(x)=(2x-8)^4(x^2+x+1)^5 is 8(2x-8)^3(x^2+x+1)^5+5(x^2+x+1)^4(2x+1)(2x-8)^4

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

derivative of

f (x) = (2 x - 8)^{4} (x^{2} + x + 1)^{5}

8 (2 x - 8)^{3} (x^{2} + x + 1)^{5} + 5 (x^{2} + x + 1)^{4} (2 x + 1) (2 x - 8)^{4}

Solution steps

\frac{d}{d x} ((2 x - 8)^{4} (x^{2} + x + 1)^{5})

Apply the Product Rule:

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

f = (2 x - 8)^{4}, g = (x^{2} + x + 1)^{5}

= \frac{d}{d x} ((2 x - 8)^{4}) (x^{2} + x + 1)^{5} + \frac{d}{d x} ((x^{2} + x + 1)^{5}) (2 x - 8)^{4}

\frac{d}{d x} ((2 x - 8)^{4}) = 8 (2 x - 8)^{3}

\frac{d}{d x} ((x^{2} + x + 1)^{5}) = 5 (x^{2} + x + 1)^{4} (2 x + 1)

= 8 (2 x - 8)^{3} (x^{2} + x + 1)^{5} + 5 (x^{2} + x + 1)^{4} (2 x + 1) (2 x - 8)^{4}

y = x^{2} - 4 x, (5, 5)

\frac{\partial}{\partial x} (x^{2} e^{yz})

\int \frac{sin ( lo g _{10} ( x ) )}{x} d x

\int \frac{1}{8} x^{2} d x

What is the derivative of f(x)=(2x-8)^4(x^2+x+1)^5 ?
The derivative of f(x)=(2x-8)^4(x^2+x+1)^5 is 8(2x-8)^3(x^2+x+1)^5+5(x^2+x+1)^4(2x+1)(2x-8)^4
What is the first derivative of f(x)=(2x-8)^4(x^2+x+1)^5 ?
The first derivative of f(x)=(2x-8)^4(x^2+x+1)^5 is 8(2x-8)^3(x^2+x+1)^5+5(x^2+x+1)^4(2x+1)(2x-8)^4