What is the derivative of x^4+8x^2-x ?

The derivative of x^4+8x^2-x is 4x^3+16x-1

What is the first derivative of x^4+8x^2-x ?

The first derivative of x^4+8x^2-x is 4x^3+16x-1

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

4 x^{3} + 16 x - 1

Solution steps

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

Apply the Sum/Difference Rule:

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

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

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

\frac{d}{d x} (8 x^{2}) = 16 x

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

= 4 x^{3} + 16 x - 1

\int_{2}^{3} (3 x^{2} + 2 x + 3) d x

y^{^{''}} + 11 y^{^{'}} + 24 y = 0

\frac{d}{d x} (\frac{\frac{d}{d x} ( x ^{5} - 5 )}{x + 1})

\int e^{- \frac{1}{4} x^{2}} x d x

\int 2 x (x^{2} + 17)^{13} d x

What is the derivative of x^4+8x^2-x ?
The derivative of x^4+8x^2-x is 4x^3+16x-1
What is the first derivative of x^4+8x^2-x ?
The first derivative of x^4+8x^2-x is 4x^3+16x-1