What is the (2sqrt(t^2+1))^' ?

The (2sqrt(t^2+1))^' is (2t)/(sqrt(t^2+1))

What is the first (2sqrt(t^2+1))^' ?

The first (2sqrt(t^2+1))^' is (2t)/(sqrt(t^2+1))

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(2 t^{2} + 1)^{^{'}}

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

Solution steps

(2 t^{2} + 1)^{'}

Take the constant out:

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

= 2 (t^{2} + 1)^{^{'}}

Apply the chain rule:

\frac{1}{2 t ^{2} + 1} (t^{2} + 1)^{'}

= \frac{1}{2 t ^{2} + 1} (t^{2} + 1)^{'}

(t^{2} + 1)^{'} = 2 t

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

Simplify

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

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

\frac{d}{d x} (14 e^{x})

\int arccot (3 x) d x

y^{^{''}} - 2 y^{^{'}} + y = 12 e^{3 t}

y^{^{'}} + y = 6

y = \frac{x}{x + 3}, (1, 2)

What is the (2sqrt(t^2+1))^' ?
The (2sqrt(t^2+1))^' is (2t)/(sqrt(t^2+1))
What is the first (2sqrt(t^2+1))^' ?
The first (2sqrt(t^2+1))^' is (2t)/(sqrt(t^2+1))