What is the derivative of 4x^3-52x^2+160x ?

The derivative of 4x^3-52x^2+160x is 12x^2-104x+160

What is the first derivative of 4x^3-52x^2+160x ?

The first derivative of 4x^3-52x^2+160x is 12x^2-104x+160

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\frac{d}{d x} (4 x^{3} - 52 x^{2} + 160 x)

12 x^{2} - 104 x + 160

Solution steps

\frac{d}{d x} (4 x^{3} - 52 x^{2} + 160 x)

Apply the Sum/Difference Rule:

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

= \frac{d}{d x} (4 x^{3}) - \frac{d}{d x} (52 x^{2}) + \frac{d}{d x} (160 x)

\frac{d}{d x} (4 x^{3}) = 12 x^{2}

\frac{d}{d x} (52 x^{2}) = 104 x

\frac{d}{d x} (160 x) = 160

= 12 x^{2} - 104 x + 160

\frac{d}{d z} (ln (z))

\int 3 x + 7 d x

\frac{( x + 1 ) d y}{d x} = x

\frac{d}{d x} (\frac{x}{x - 3} - arcsec (2 x))

x \to 0 lim (\frac{x}{tan ( x )})

What is the derivative of 4x^3-52x^2+160x ?
The derivative of 4x^3-52x^2+160x is 12x^2-104x+160
What is the first derivative of 4x^3-52x^2+160x ?
The first derivative of 4x^3-52x^2+160x is 12x^2-104x+160