What is the derivative of (4x^2+2x-5(x^3+7x+4)) ?

The derivative of (4x^2+2x-5(x^3+7x+4)) is 20x^4+8x^3+69x^2+60x-27

What is the first derivative of (4x^2+2x-5(x^3+7x+4)) ?

The first derivative of (4x^2+2x-5(x^3+7x+4)) is 20x^4+8x^3+69x^2+60x-27

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\frac{d}{d x} ((4 x^{2} + 2 x - 5) (x^{3} + 7 x + 4))

20 x^{4} + 8 x^{3} + 69 x^{2} + 60 x - 27

Solution steps

\frac{d}{d x} ((4 x^{2} + 2 x - 5) (x^{3} + 7 x + 4))

Apply the Product Rule:

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

f = 4 x^{2} + 2 x - 5, g = x^{3} + 7 x + 4

= \frac{d}{d x} (4 x^{2} + 2 x - 5) (x^{3} + 7 x + 4) + \frac{d}{d x} (x^{3} + 7 x + 4) (4 x^{2} + 2 x - 5)

\frac{d}{d x} (4 x^{2} + 2 x - 5) = 8 x + 2

\frac{d}{d x} (x^{3} + 7 x + 4) = 3 x^{2} + 7

= (8 x + 2) (x^{3} + 7 x + 4) + (3 x^{2} + 7) (4 x^{2} + 2 x - 5)

Simplify

(8 x + 2) (x^{3} + 7 x + 4) + (3 x^{2} + 7) (4 x^{2} + 2 x - 5) : 20 x^{4} + 8 x^{3} + 69 x^{2} + 60 x - 27

= 20 x^{4} + 8 x^{3} + 69 x^{2} + 60 x - 27

d er i v a t i v e \frac{4 u ^{2}}{( u ^{2} + u ) ^{3}}

d er i v a t i v e f (x) = a cos (2 x) + b sin (2 x)

\int_{1}^{\infty} 15 e^{- 5 x} d x

\frac{d}{d x} (cos (2 x) + x^{2})

d er i v a t i v e f (x) = (1 - x^{3}) + (1 - x^{3})

What is the derivative of (4x^2+2x-5(x^3+7x+4)) ?
The derivative of (4x^2+2x-5(x^3+7x+4)) is 20x^4+8x^3+69x^2+60x-27
What is the first derivative of (4x^2+2x-5(x^3+7x+4)) ?
The first derivative of (4x^2+2x-5(x^3+7x+4)) is 20x^4+8x^3+69x^2+60x-27