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

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

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

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

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derivative of

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

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

Solution steps

\frac{d}{d x} ((2 x - 3)^{4} (x^{2} + x + 1)^{5})

Apply the Product Rule:

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

f = (2 x - 3)^{4}, g = (x^{2} + x + 1)^{5}

= \frac{d}{d x} ((2 x - 3)^{4}) (x^{2} + x + 1)^{5} + \frac{d}{d x} ((x^{2} + x + 1)^{5}) (2 x - 3)^{4}

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

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

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

\int \frac{1}{x ^{2} + 6 x + 13} d x

\int_{1}^{4} 3 e^{x} d x

\frac{d}{d x} (tan (π))

x = 1, y = 6 x (2 - x^{2})^{2}

x \to - 4 - lim (\frac{- 4 x ^{3}}{4 + x})

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