What is the derivative of x^4-8x^2+17 ?

The derivative of x^4-8x^2+17 is 4x^3-16x

What is the first derivative of x^4-8x^2+17 ?

The first derivative of x^4-8x^2+17 is 4x^3-16x

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\frac{d}{d x} (x^{4} - 8 x^{2} + 17)

4 x^{3} - 16 x

Solution steps

\frac{d}{d x} (x^{4} - 8 x^{2} + 17)

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}{d x} (17)

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

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

\frac{d}{d x} (17) = 0

= 4 x^{3} - 16 x + 0

Simplify

= 4 x^{3} - 16 x

\frac{d}{d x} (3 csc (x) + 6 cos (x))

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

\frac{1}{5} x^{5}

\frac{\partial}{\partial p ( r , t )} (p (r, t) e^{r t})

\int sin (x) sin (n x) d x

What is the derivative of x^4-8x^2+17 ?
The derivative of x^4-8x^2+17 is 4x^3-16x
What is the first derivative of x^4-8x^2+17 ?
The first derivative of x^4-8x^2+17 is 4x^3-16x