What is the derivative of p(x)=x^4-3x^3+10x^2-6x-20 ?

The derivative of p(x)=x^4-3x^3+10x^2-6x-20 is 4x^3-9x^2+20x-6

What is the first derivative of p(x)=x^4-3x^3+10x^2-6x-20 ?

The first derivative of p(x)=x^4-3x^3+10x^2-6x-20 is 4x^3-9x^2+20x-6

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derivative of

p (x) = x^{4} - 3 x^{3} + 10 x^{2} - 6 x - 20

4 x^{3} - 9 x^{2} + 20 x - 6

Solution steps

\frac{d}{d x} (x^{4} - 3 x^{3} + 10 x^{2} - 6 x - 20)

Apply the Sum/Difference Rule:

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

= \frac{d}{d x} (x^{4}) - \frac{d}{d x} (3 x^{3}) + \frac{d}{d x} (10 x^{2}) - \frac{d}{d x} (6 x) - \frac{d}{d x} (20)

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

\frac{d}{d x} (3 x^{3}) = 9 x^{2}

\frac{d}{d x} (10 x^{2}) = 20 x

\frac{d}{d x} (6 x) = 6

\frac{d}{d x} (20) = 0

= 4 x^{3} - 9 x^{2} + 20 x - 6 - 0

Simplify

= 4 x^{3} - 9 x^{2} + 20 x - 6

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

f (x) = 9 e^{x} + x, at x = 0

\int - e^{- 2 t} d t

\frac{\partial}{\partial x} (\frac{1}{2 π t} e^{- \frac{( x - y ) ^{2}}{2 t}})

x \to 1 lim (x^{2} - 2)

What is the derivative of p(x)=x^4-3x^3+10x^2-6x-20 ?
The derivative of p(x)=x^4-3x^3+10x^2-6x-20 is 4x^3-9x^2+20x-6
What is the first derivative of p(x)=x^4-3x^3+10x^2-6x-20 ?
The first derivative of p(x)=x^4-3x^3+10x^2-6x-20 is 4x^3-9x^2+20x-6