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

x \to 1 lim (\frac{x}{x ^{2} - x})

\frac{4 u ^{2}}{( u ^{2} + u ) ^{3}}

f (x) = a cos (2 x) + b sin (2 x)

\frac{\partial}{\partial y} (x^{3} e^{- 2 y})

\int \frac{900}{( 2 q - 5 ) ^{2}} d q

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