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

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

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

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

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derivative of

y = 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 x^{3} - \frac{2}{3} x^{2} + 7 x d x

n = 0 \sum \infty \frac{8}{4 ^{n}}

\frac{\partial}{\partial x} (- sin (x^{2} y) x^{2})

\int t \cdot \frac{1}{e ^{t}} d t

\int (- 2 sec (x) tan (x) - 3 sec^{2} (x)) d x

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