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

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

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

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

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derivative of

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

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

Solution steps

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

Apply the Product Rule:

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

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

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

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

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

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

\frac{\partial}{\partial x} (\frac{y}{2 - x})

f (x) = \frac{4 x - 1}{x + 5}

f (x) = (5 x - x^{2}) (5 - x - x^{2})

\frac{\partial}{\partial t} (t^{2} x - t)

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