What is the d/(dt)(6arctan(t)) ?

The d/(dt)(6arctan(t)) is 6/(t^2+1)

What is the first d/(dt)(6arctan(t)) ?

The first d/(dt)(6arctan(t)) is 6/(t^2+1)

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\frac{d}{d t} (6 arctan (t))

\frac{6}{t ^{2} + 1}

Solution steps

\frac{d}{d t} (6 arctan (t))

Take the constant out:

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

= 6 \frac{d}{d t} (arctan (t))

Apply the common derivative:

\frac{d}{d t} (arctan (t)) = \frac{1}{t ^{2} + 1}

= 6 \cdot \frac{1}{t ^{2} + 1}

Simplify

6 \cdot \frac{1}{t ^{2} + 1} : \frac{6}{t ^{2} + 1}

= \frac{6}{t ^{2} + 1}

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\frac{\partial}{\partial z} (x e^{- z})

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What is the d/(dt)(6arctan(t)) ?
The d/(dt)(6arctan(t)) is 6/(t^2+1)
What is the first d/(dt)(6arctan(t)) ?
The first d/(dt)(6arctan(t)) is 6/(t^2+1)