What is the derivative of (-pix^2-4x^2+64x/8) ?

The derivative of (-pix^2-4x^2+64x/8) is (-pix-4x+32)/4

What is the first derivative of (-pix^2-4x^2+64x/8) ?

The first derivative of (-pix^2-4x^2+64x/8) is (-pix-4x+32)/4

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

\frac{d}{d x} (\frac{- π x ^{2} - 4 x ^{2} + 64 x}{8})

\frac{- π x - 4 x + 32}{4}

Solution steps

\frac{d}{d x} (\frac{- π x ^{2} - 4 x ^{2} + 64 x}{8})

Take the constant out:

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

= \frac{1}{8} \frac{d}{d x} (- π x^{2} - 4 x^{2} + 64 x)

Apply the Sum/Difference Rule:

(f \pm g)^{'} = f^{'} \pm g^{'}

= \frac{1}{8} (- \frac{d}{d x} (π x^{2}) - \frac{d}{d x} (4 x^{2}) + \frac{d}{d x} (64 x))

\frac{d}{d x} (π x^{2}) = 2 π x

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

\frac{d}{d x} (64 x) = 64

= \frac{1}{8} (- 2 π x - 8 x + 64)

Simplify

\frac{1}{8} (- 2 π x - 8 x + 64) : \frac{- π x - 4 x + 32}{4}

= \frac{- π x - 4 x + 32}{4}

What is the derivative of (-pix^2-4x^2+64x/8) ?
The derivative of (-pix^2-4x^2+64x/8) is (-pix-4x+32)/4
What is the first derivative of (-pix^2-4x^2+64x/8) ?
The first derivative of (-pix^2-4x^2+64x/8) is (-pix-4x+32)/4