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

The derivative of (3x^4-x+1)(-x^5+4) is -27x^8+6x^5-5x^4+48x^3-4

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

The first derivative of (3x^4-x+1)(-x^5+4) is -27x^8+6x^5-5x^4+48x^3-4

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derivative of

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

- 27 x^{8} + 6 x^{5} - 5 x^{4} + 48 x^{3} - 4

Solution steps

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

Apply the Product Rule:

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

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

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

\frac{d}{d x} (3 x^{4} - x + 1) = 12 x^{3} - 1

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

= (12 x^{3} - 1) (- x^{5} + 4) + (- 5 x^{4}) (3 x^{4} - x + 1)

Simplify

(12 x^{3} - 1) (- x^{5} + 4) + (- 5 x^{4}) (3 x^{4} - x + 1) : - 27 x^{8} + 6 x^{5} - 5 x^{4} + 48 x^{3} - 4

= - 27 x^{8} + 6 x^{5} - 5 x^{4} + 48 x^{3} - 4

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

f (t) = sin (2 t)

\int - \frac{1}{( t + 1 ) ^{2}} d t

\frac{d}{d x} (x^{2} + 4 x (\frac{36.5}{x ^{2}}))

f (x) = x + x, (4, 6)

What is the derivative of (3x^4-x+1)(-x^5+4) ?
The derivative of (3x^4-x+1)(-x^5+4) is -27x^8+6x^5-5x^4+48x^3-4
What is the first derivative of (3x^4-x+1)(-x^5+4) ?
The first derivative of (3x^4-x+1)(-x^5+4) is -27x^8+6x^5-5x^4+48x^3-4