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

The derivative of (-x^2+4x+5(4x^4-3)) is -24x^5+80x^4+80x^3+6x-12

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

The first derivative of (-x^2+4x+5(4x^4-3)) is -24x^5+80x^4+80x^3+6x-12

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

- 24 x^{5} + 80 x^{4} + 80 x^{3} + 6 x - 12

Solution steps

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

Apply the Product Rule:

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

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

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

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

\frac{d}{d x} (4 x^{4} - 3) = 16 x^{3}

= (- 2 x + 4) (4 x^{4} - 3) + 16 x^{3} (- x^{2} + 4 x + 5)

Simplify

(- 2 x + 4) (4 x^{4} - 3) + 16 x^{3} (- x^{2} + 4 x + 5) : - 24 x^{5} + 80 x^{4} + 80 x^{3} + 6 x - 12

= - 24 x^{5} + 80 x^{4} + 80 x^{3} + 6 x - 12

\frac{d x}{d t} = 3 x t^{3}

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

x \to 4 + lim (2 x - 9)

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

\int 1 + \frac{9}{x ^{2} - 4} d x

What is the derivative of (-x^2+4x+5(4x^4-3)) ?
The derivative of (-x^2+4x+5(4x^4-3)) is -24x^5+80x^4+80x^3+6x-12
What is the first derivative of (-x^2+4x+5(4x^4-3)) ?
The first derivative of (-x^2+4x+5(4x^4-3)) is -24x^5+80x^4+80x^3+6x-12