What is the d/(dt)(sqrt(t^2+1)) ?

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

What is the first d/(dt)(sqrt(t^2+1)) ?

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

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

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

Solution steps

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

Apply the chain rule:

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

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

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

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

Simplify

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

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

\frac{d y}{d t} = \frac{t ^{2}}{y ( 1 + t ^{3} )}

\int_{3}^{4} \frac{x}{8} d x

\frac{\partial}{\partial r} (π r^{2} h)

\frac{11}{x}

\int \frac{1}{9 - 2 x} d x

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