What is the derivative of 1-(1-x/(24)^3) ?

The derivative of 1-(1-x/(24)^3) is (x^2)/(4608)-x/(96)+1/8

What is the first derivative of 1-(1-x/(24)^3) ?

The first derivative of 1-(1-x/(24)^3) is (x^2)/(4608)-x/(96)+1/8

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\frac{d}{d x} (1 - (1 - \frac{x}{24})^{3})

\frac{x ^{2}}{4608} - \frac{x}{96} + \frac{1}{8}

Solution steps

\frac{d}{d x} (1 - (1 - \frac{x}{24})^{3})

Simplify

1 - (1 - \frac{x}{24})^{3} : \frac{x ^{3}}{13824} - \frac{x ^{2}}{192} + \frac{x}{8}

= \frac{d}{d x} (\frac{x ^{3}}{13824} - \frac{x ^{2}}{192} + \frac{x}{8})

Apply the Sum/Difference Rule:

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

= \frac{d}{d x} (\frac{x ^{3}}{13824}) - \frac{d}{d x} (\frac{x ^{2}}{192}) + \frac{d}{d x} (\frac{x}{8})

\frac{d}{d x} (\frac{x ^{3}}{13824}) = \frac{x ^{2}}{4608}

\frac{d}{d x} (\frac{x ^{2}}{192}) = \frac{x}{96}

\frac{d}{d x} (\frac{x}{8}) = \frac{1}{8}

= \frac{x ^{2}}{4608} - \frac{x}{96} + \frac{1}{8}

y^{^{'}} + 3 y = 6

f (x) = (x - 5)^{6} (x + 3)^{2}

\frac{d}{d x} (sin^{\frac{1}{2}} (2 x))

\int - 8 (y)^{2} (1) d y

x = y^{2} + 14 y + 16, x = 7 y + 6

What is the derivative of 1-(1-x/(24)^3) ?
The derivative of 1-(1-x/(24)^3) is (x^2)/(4608)-x/(96)+1/8
What is the first derivative of 1-(1-x/(24)^3) ?
The first derivative of 1-(1-x/(24)^3) is (x^2)/(4608)-x/(96)+1/8