What is the derivative of 1/5 x^5-3x^4+9x^2+1 ?

The derivative of 1/5 x^5-3x^4+9x^2+1 is x^4-12x^3+18x

What is the first derivative of 1/5 x^5-3x^4+9x^2+1 ?

The first derivative of 1/5 x^5-3x^4+9x^2+1 is x^4-12x^3+18x

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\frac{d}{d x} (\frac{1}{5} x^{5} - 3 x^{4} + 9 x^{2} + 1)

x^{4} - 12 x^{3} + 18 x

Solution steps

\frac{d}{d x} (\frac{1}{5} x^{5} - 3 x^{4} + 9 x^{2} + 1)

Simplify

\frac{1}{5} x^{5} - 3 x^{4} + 9 x^{2} + 1 : \frac{x ^{5}}{5} - 3 x^{4} + 9 x^{2} + 1

= \frac{d}{d x} (\frac{x ^{5}}{5} - 3 x^{4} + 9 x^{2} + 1)

Apply the Sum/Difference Rule:

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

= \frac{d}{d x} (\frac{x ^{5}}{5}) - \frac{d}{d x} (3 x^{4}) + \frac{d}{d x} (9 x^{2}) + \frac{d}{d x} (1)

\frac{d}{d x} (\frac{x ^{5}}{5}) = x^{4}

\frac{d}{d x} (3 x^{4}) = 12 x^{3}

\frac{d}{d x} (9 x^{2}) = 18 x

\frac{d}{d x} (1) = 0

= x^{4} - 12 x^{3} + 18 x + 0

Simplify

= x^{4} - 12 x^{3} + 18 x

\frac{\partial}{\partial x} (cos (2 x + y) - z)

\frac{d y}{d x} = \frac{6 y}{x}

x \to 0 lim (\frac{sin ( 6 x )}{2})

\int \frac{x}{3 x ^{2} - 5} d x

\frac{d y}{d x} = 3 x y^{2}

What is the derivative of 1/5 x^5-3x^4+9x^2+1 ?
The derivative of 1/5 x^5-3x^4+9x^2+1 is x^4-12x^3+18x
What is the first derivative of 1/5 x^5-3x^4+9x^2+1 ?
The first derivative of 1/5 x^5-3x^4+9x^2+1 is x^4-12x^3+18x