What is the derivative of 1-(x^2/(16)) ?

The derivative of 1-(x^2/(16)) is -x/8

What is the first derivative of 1-(x^2/(16)) ?

The first derivative of 1-(x^2/(16)) is -x/8

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\frac{d}{d x} (1 - \frac{x ^{2}}{16})

- \frac{x}{8}

Solution steps

\frac{d}{d x} (1 - \frac{x ^{2}}{16})

Apply the Sum/Difference Rule:

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

= \frac{d}{d x} (1) - \frac{d}{d x} (\frac{x ^{2}}{16})

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

\frac{d}{d x} (\frac{x ^{2}}{16}) = \frac{x}{8}

= 0 - \frac{x}{8}

Simplify

= - \frac{x}{8}

y^{2} d x + x (x - y) d y = 0

\int_{- 1}^{1} 2 x d x

2 t^{2} - 3 t + 3

x \to \infty lim (5 x^{4})

\frac{\frac{1}{s + 1} + \frac{3}{2}}{s + 2}

What is the derivative of 1-(x^2/(16)) ?
The derivative of 1-(x^2/(16)) is -x/8
What is the first derivative of 1-(x^2/(16)) ?
The first derivative of 1-(x^2/(16)) is -x/8