What is the first derivative of 2t^2 ?

The first derivative of 2t^2 is 4t

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derivative of 2t^2

derivative of

2 t^{2}

4 t

Solution steps

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

Take the constant out:

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

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

Apply the Power Rule:

\frac{d}{d x} (x^{a}) = a \cdot x^{a - 1}

= 2 \cdot 2 t^{2 - 1}

Simplify

= 4 t

\frac{d}{d x} (arctan (\frac{9}{x}))

f (x) = 5 x + 6, (6, 6)

\int (tan (3 x) + cot (3 x))^{2} d x

n = 0 \sum \infty \frac{5 ^{n}}{3 ^{n}}

\frac{s}{s - 1}