What is the derivative of (1-t^2)/4 ?

The derivative of (1-t^2)/4 is -t/2

What is the first derivative of (1-t^2)/4 ?

The first derivative of (1-t^2)/4 is -t/2

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derivative of

\frac{1 - t ^{2}}{4}

- \frac{t}{2}

Solution steps

\frac{d}{d t} (\frac{1 - t ^{2}}{4})

Take the constant out:

(a \cdot f)^{'} = a \cdot f^{'}

= \frac{1}{4} \frac{d}{d t} (1 - t^{2})

Apply the Sum/Difference Rule:

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

= \frac{1}{4} (\frac{d}{d t} (1) - \frac{d}{d t} (t^{2}))

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

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

= \frac{1}{4} (0 - 2 t)

Simplify

\frac{1}{4} (0 - 2 t) : - \frac{t}{2}

= - \frac{t}{2}

\frac{d}{d s} (\frac{( 13 ( s + 2 ) ( s + 5 ) )}{( s ( s + 3 ) )})

q (t) = 6 t^{2} + 5 t + 4

\frac{5}{( 1 + 3 x )}

\frac{d}{d t} (- 38 t + 220)

(x^{2} + 9) (x^{3} + 5)

What is the derivative of (1-t^2)/4 ?
The derivative of (1-t^2)/4 is -t/2
What is the first derivative of (1-t^2)/4 ?
The first derivative of (1-t^2)/4 is -t/2