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

The derivative of (4x^4-x+3)(-x^5+5) is -36x^8+6x^5-15x^4+80x^3-5

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

The first derivative of (4x^4-x+3)(-x^5+5) is -36x^8+6x^5-15x^4+80x^3-5

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derivative of

(4 x^{4} - x + 3) (- x^{5} + 5)

- 36 x^{8} + 6 x^{5} - 15 x^{4} + 80 x^{3} - 5

Solution steps

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

Apply the Product Rule:

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

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

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

\frac{d}{d x} (4 x^{4} - x + 3) = 16 x^{3} - 1

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

= (16 x^{3} - 1) (- x^{5} + 5) + (- 5 x^{4}) (4 x^{4} - x + 3)

Simplify

(16 x^{3} - 1) (- x^{5} + 5) + (- 5 x^{4}) (4 x^{4} - x + 3) : - 36 x^{8} + 6 x^{5} - 15 x^{4} + 80 x^{3} - 5

= - 36 x^{8} + 6 x^{5} - 15 x^{4} + 80 x^{3} - 5

\frac{\partial}{\partial x} (\frac{m}{v})

\int e^{s i n (a)} cos (a)

x = y^{2} - 2, y = x

\int \frac{5 ln ( x + 8 )}{x ^{2}} d x

sin (x) - x

What is the derivative of (4x^4-x+3)(-x^5+5) ?
The derivative of (4x^4-x+3)(-x^5+5) is -36x^8+6x^5-15x^4+80x^3-5
What is the first derivative of (4x^4-x+3)(-x^5+5) ?
The first derivative of (4x^4-x+3)(-x^5+5) is -36x^8+6x^5-15x^4+80x^3-5