What is the derivative of-1/6 x^3-1/2 x^2+x+1+ke^k ?

The derivative of-1/6 x^3-1/2 x^2+x+1+ke^k is -(x^2)/2-x+1

What is the first derivative of-1/6 x^3-1/2 x^2+x+1+ke^k ?

The first derivative of-1/6 x^3-1/2 x^2+x+1+ke^k is -(x^2)/2-x+1

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\frac{d}{d x} (- \frac{1}{6} x^{3} - \frac{1}{2} x^{2} + x + 1 + k e^{k})

- \frac{x ^{2}}{2} - x + 1

Solution steps

\frac{d}{d x} (- \frac{1}{6} x^{3} - \frac{1}{2} x^{2} + x + 1 + k e^{k})

Simplify

- \frac{1}{6} x^{3} - \frac{1}{2} x^{2} + x + 1 + k e^{k} : - \frac{x ^{3}}{6} - \frac{x ^{2}}{2} + x + 1 + k e^{k}

= \frac{d}{d x} (- \frac{x ^{3}}{6} - \frac{x ^{2}}{2} + x + 1 + k e^{k})

Apply the Sum/Difference Rule:

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

= - \frac{d}{d x} (\frac{x ^{3}}{6}) - \frac{d}{d x} (\frac{x ^{2}}{2}) + \frac{d x}{d x} + \frac{d}{d x} (1) + \frac{d}{d x} (k e^{k})

\frac{d}{d x} (\frac{x ^{3}}{6}) = \frac{x ^{2}}{2}

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

\frac{d x}{d x} = 1

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

\frac{d}{d x} (k e^{k}) = 0

= - \frac{x ^{2}}{2} - x + 1 + 0 + 0

Simplify

= - \frac{x ^{2}}{2} - x + 1

\int 8 sec^{4} (θ) tan^{2} (θ) d θ

\int 3 - 2 x d x

\frac{\partial}{\partial x} (x y + \frac{1}{x} + \frac{8}{y})

n = 1 \sum \infty \frac{1}{( 3 + 2 n ) ^{2}}

What is the derivative of-1/6 x^3-1/2 x^2+x+1+ke^k ?
The derivative of-1/6 x^3-1/2 x^2+x+1+ke^k is -(x^2)/2-x+1
What is the first derivative of-1/6 x^3-1/2 x^2+x+1+ke^k ?
The first derivative of-1/6 x^3-1/2 x^2+x+1+ke^k is -(x^2)/2-x+1