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{x cos ( x )}{1 + x ^{2}}

1 + w^{2}

2 (x sin (x) + cos (x))

\frac{1}{3 x ^{2} + 2}

\frac{( cos ( x ) - cos ^{3} ( x ) )}{x ^{2}}

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