What is the derivative of (x^4)/4-(3x^2)/4+x-2 ?

The derivative of (x^4)/4-(3x^2)/4+x-2 is (x(2x^2-3))/2+1

What is the first derivative of (x^4)/4-(3x^2)/4+x-2 ?

The first derivative of (x^4)/4-(3x^2)/4+x-2 is (x(2x^2-3))/2+1

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derivative of

\frac{x ^{4}}{4} - \frac{3 x ^{2}}{4} + x - 2

\frac{x ( 2 x ^{2} - 3 )}{2} + 1

Solution steps

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

Simplify

\frac{x ^{4}}{4} - \frac{3 x ^{2}}{4} + x - 2 : \frac{x ^{4} - 3 x ^{2}}{4} + x - 2

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

Apply the Sum/Difference Rule:

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

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

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

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

\frac{d}{d x} (2) = 0

= \frac{x ( 2 x ^{2} - 3 )}{2} + 1 - 0

Simplify

= \frac{x ( 2 x ^{2} - 3 )}{2} + 1

\frac{d}{d x} (\frac{x + 2}{ln ( x )})

f (x) = x^{2} - 7 x + 7, (6, 1)

y^{^{''}} + 3 y^{^{'}} + 2 y = 20 e^{3 t}

\frac{3}{s ( s ^{3} + 2 s ^{2} + 2 s + 1 )}

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

What is the derivative of (x^4)/4-(3x^2)/4+x-2 ?
The derivative of (x^4)/4-(3x^2)/4+x-2 is (x(2x^2-3))/2+1
What is the first derivative of (x^4)/4-(3x^2)/4+x-2 ?
The first derivative of (x^4)/4-(3x^2)/4+x-2 is (x(2x^2-3))/2+1