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

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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^{3} - 10 x^{2} + 21 x, y = - x^{3} + 10 x^{2} - 21 x

e^{\frac{3}{x}}

y^{^{''}} - y = \frac{2 e ^{- x}}{( 1 + e ^{- 2 x} ) ^{2}}

\frac{d y}{d x} + \frac{x y}{x ^{2} + 1} = \frac{x}{x ^{2} + 1}

n \to \infty lim (- 3^{- n})

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