What is the d/(dt)((t^2)/(sqrt(t^3+1))) ?

The d/(dt)((t^2)/(sqrt(t^3+1))) is (t^4+4t)/(2(t^3+1)sqrt(t^3+1))

What is the first d/(dt)((t^2)/(sqrt(t^3+1))) ?

The first d/(dt)((t^2)/(sqrt(t^3+1))) is (t^4+4t)/(2(t^3+1)sqrt(t^3+1))

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

\frac{t ^{4} + 4 t}{2 ( t ^{3} + 1 ) t ^{3} + 1}

Solution steps

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

Apply the Quotient Rule:

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

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

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

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

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

Simplify

\frac{2 t t ^{3} + 1 - \frac{3 t ^{2}}{2 t ^{3} + 1} t ^{2}}{( t ^{3} + 1 ) ^{2}} : \frac{t ^{4} + 4 t}{2 ( t ^{3} + 1 ) t ^{3} + 1}

= \frac{t ^{4} + 4 t}{2 ( t ^{3} + 1 ) t ^{3} + 1}

- x^{2} sin (x) + 4 x cos (x) + 2 sin (x)

\int - 8 ln (x^{2} - 1) d x

\int_{- 2}^{3} (x^{2} - 12) - (x - 6) d x

x \to 1 lim (x - 1)

x \to 1 lim (3 e^{x - 1})

What is the d/(dt)((t^2)/(sqrt(t^3+1))) ?
The d/(dt)((t^2)/(sqrt(t^3+1))) is (t^4+4t)/(2(t^3+1)sqrt(t^3+1))
What is the first d/(dt)((t^2)/(sqrt(t^3+1))) ?
The first d/(dt)((t^2)/(sqrt(t^3+1))) is (t^4+4t)/(2(t^3+1)sqrt(t^3+1))