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

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

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

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

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

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

Solution steps

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

Apply the Quotient Rule:

(\frac{f}{g})^{'} = \frac{f ^{'} \cdot g - g ^{'} \cdot f}{g ^{2}}

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

\frac{d t}{d t} = 1

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

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

Simplify

\frac{1 \cdot 1 + t ^{2} - \frac{t}{1 + t ^{2}} t}{( 1 + t ^{2} ) ^{2}} : \frac{1}{( 1 + t ^{2} ) 1 + t ^{2}}

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

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