What is the derivative of f(x)=-64x(-16x^2+36) ?

The derivative of f(x)=-64x(-16x^2+36) is -64(-48x^2+36)

What is the first derivative of f(x)=-64x(-16x^2+36) ?

The first derivative of f(x)=-64x(-16x^2+36) is -64(-48x^2+36)

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derivative of

f (x) = - 64 x (- 16 x^{2} + 36)

- 64 (- 48 x^{2} + 36)

Solution steps

\frac{d}{d x} (- 64 x (- 16 x^{2} + 36))

Take the constant out:

(a \cdot f)^{'} = a \cdot f^{'}

= - 64 \frac{d}{d x} (x (- 16 x^{2} + 36))

Apply the Product Rule:

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

f = x, g = - 16 x^{2} + 36

= - 64 (\frac{d x}{d x} (- 16 x^{2} + 36) + \frac{d}{d x} (- 16 x^{2} + 36) x)

\frac{d x}{d x} = 1

\frac{d}{d x} (- 16 x^{2} + 36) = - 32 x

= - 64 (1 \cdot (- 16 x^{2} + 36) + (- 32 x) x)

Simplify

- 64 (1 \cdot (- 16 x^{2} + 36) + (- 32 x) x) : - 64 (- 48 x^{2} + 36)

= - 64 (- 48 x^{2} + 36)

(x^{2} - y^{2}) d x - 2 x y d y = 0

\frac{\partial}{\partial x} ((\frac{x}{y}) e^{x y})

\int 5 x cos (2 x) d x

\frac{d y}{d x}, y = 7 x^{- 4}

\int 2 x cos (2 x) d x

What is the derivative of f(x)=-64x(-16x^2+36) ?
The derivative of f(x)=-64x(-16x^2+36) is -64(-48x^2+36)
What is the first derivative of f(x)=-64x(-16x^2+36) ?
The first derivative of f(x)=-64x(-16x^2+36) is -64(-48x^2+36)