What is the derivative of x^4+8x^3+6x^2-40x+25 ?

The derivative of x^4+8x^3+6x^2-40x+25 is 4x^3+24x^2+12x-40

What is the first derivative of x^4+8x^3+6x^2-40x+25 ?

The first derivative of x^4+8x^3+6x^2-40x+25 is 4x^3+24x^2+12x-40

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\frac{d}{d x} (x^{4} + 8 x^{3} + 6 x^{2} - 40 x + 25)

4 x^{3} + 24 x^{2} + 12 x - 40

Solution steps

\frac{d}{d x} (x^{4} + 8 x^{3} + 6 x^{2} - 40 x + 25)

Apply the Sum/Difference Rule:

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

= \frac{d}{d x} (x^{4}) + \frac{d}{d x} (8 x^{3}) + \frac{d}{d x} (6 x^{2}) - \frac{d}{d x} (40 x) + \frac{d}{d x} (25)

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

\frac{d}{d x} (8 x^{3}) = 24 x^{2}

\frac{d}{d x} (6 x^{2}) = 12 x

\frac{d}{d x} (40 x) = 40

\frac{d}{d x} (25) = 0

= 4 x^{3} + 24 x^{2} + 12 x - 40 + 0

Simplify

= 4 x^{3} + 24 x^{2} + 12 x - 40

x y^{^{'}} + 2 y = - 3 sin (x)

\int \frac{x ^{3}}{64 - x ^{2}} d x

\int \frac{1}{x ln ( x ^{5} )} d x

\int e^{x^{2} + 1} d x

\frac{d}{d x} ((8^{\frac{2}{3}} - x^{\frac{2}{3}})^{\frac{3}{2}})

What is the derivative of x^4+8x^3+6x^2-40x+25 ?
The derivative of x^4+8x^3+6x^2-40x+25 is 4x^3+24x^2+12x-40
What is the first derivative of x^4+8x^3+6x^2-40x+25 ?
The first derivative of x^4+8x^3+6x^2-40x+25 is 4x^3+24x^2+12x-40