What is the derivative of (7x+1(x^4-x^3-9x)) ?

The derivative of (7x+1(x^4-x^3-9x)) is 35x^4-24x^3-3x^2-126x-9

What is the first derivative of (7x+1(x^4-x^3-9x)) ?

The first derivative of (7x+1(x^4-x^3-9x)) is 35x^4-24x^3-3x^2-126x-9

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\frac{d}{d x} ((7 x + 1) (x^{4} - x^{3} - 9 x))

35 x^{4} - 24 x^{3} - 3 x^{2} - 126 x - 9

Solution steps

\frac{d}{d x} ((7 x + 1) (x^{4} - x^{3} - 9 x))

Apply the Product Rule:

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

f = 7 x + 1, g = x^{4} - x^{3} - 9 x

= \frac{d}{d x} (7 x + 1) (x^{4} - x^{3} - 9 x) + \frac{d}{d x} (x^{4} - x^{3} - 9 x) (7 x + 1)

\frac{d}{d x} (7 x + 1) = 7

\frac{d}{d x} (x^{4} - x^{3} - 9 x) = 4 x^{3} - 3 x^{2} - 9

= 7 (x^{4} - x^{3} - 9 x) + (4 x^{3} - 3 x^{2} - 9) (7 x + 1)

Simplify

7 (x^{4} - x^{3} - 9 x) + (4 x^{3} - 3 x^{2} - 9) (7 x + 1) : 35 x^{4} - 24 x^{3} - 3 x^{2} - 126 x - 9

= 35 x^{4} - 24 x^{3} - 3 x^{2} - 126 x - 9

y^{^{''}} + 4 y^{^{'}} + 3 y = 2 x

\frac{d}{d x} (\frac{e ^{x}}{2 x ^{2}})

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

\frac{d}{d x} (16 + x)

x \to 0 lim (2^{- x})

What is the derivative of (7x+1(x^4-x^3-9x)) ?
The derivative of (7x+1(x^4-x^3-9x)) is 35x^4-24x^3-3x^2-126x-9
What is the first derivative of (7x+1(x^4-x^3-9x)) ?
The first derivative of (7x+1(x^4-x^3-9x)) is 35x^4-24x^3-3x^2-126x-9