What is the derivative of (x^2-16x+64)(x^2+64x+4) ?

The derivative of (x^2-16x+64)(x^2+64x+4) is 4x^3+144x^2-1912x+4032

What is the first derivative of (x^2-16x+64)(x^2+64x+4) ?

The first derivative of (x^2-16x+64)(x^2+64x+4) is 4x^3+144x^2-1912x+4032

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derivative of

(x^{2} - 16 x + 64) (x^{2} + 64 x + 4)

4 x^{3} + 144 x^{2} - 1912 x + 4032

Solution steps

\frac{d}{d x} ((x^{2} - 16 x + 64) (x^{2} + 64 x + 4))

Apply the Product Rule:

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

f = x^{2} - 16 x + 64, g = x^{2} + 64 x + 4

= \frac{d}{d x} (x^{2} - 16 x + 64) (x^{2} + 64 x + 4) + \frac{d}{d x} (x^{2} + 64 x + 4) (x^{2} - 16 x + 64)

\frac{d}{d x} (x^{2} - 16 x + 64) = 2 x - 16

\frac{d}{d x} (x^{2} + 64 x + 4) = 2 x + 64

= (2 x - 16) (x^{2} + 64 x + 4) + (2 x + 64) (x^{2} - 16 x + 64)

Simplify

(2 x - 16) (x^{2} + 64 x + 4) + (2 x + 64) (x^{2} - 16 x + 64) : 4 x^{3} + 144 x^{2} - 1912 x + 4032

= 4 x^{3} + 144 x^{2} - 1912 x + 4032

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

\frac{d y}{d x} = (\frac{x - y}{x})

(cos (5 x))^{^{'}}

f (x) = \frac{6 x}{x + 5}

8 x e^{- k x}

What is the derivative of (x^2-16x+64)(x^2+64x+4) ?
The derivative of (x^2-16x+64)(x^2+64x+4) is 4x^3+144x^2-1912x+4032
What is the first derivative of (x^2-16x+64)(x^2+64x+4) ?
The first derivative of (x^2-16x+64)(x^2+64x+4) is 4x^3+144x^2-1912x+4032