What is the derivative of f(t)=(t^2)/(sqrt(t^3+1)) ?

The derivative of f(t)=(t^2)/(sqrt(t^3+1)) is (t^4+4t)/(2(t^3+1)sqrt(t^3+1))

What is the first derivative of f(t)=(t^2)/(sqrt(t^3+1)) ?

The first derivative of f(t)=(t^2)/(sqrt(t^3+1)) is (t^4+4t)/(2(t^3+1)sqrt(t^3+1))

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derivative of

f (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}

What is the derivative of f(t)=(t^2)/(sqrt(t^3+1)) ?
The derivative of f(t)=(t^2)/(sqrt(t^3+1)) is (t^4+4t)/(2(t^3+1)sqrt(t^3+1))
What is the first derivative of f(t)=(t^2)/(sqrt(t^3+1)) ?
The first derivative of f(t)=(t^2)/(sqrt(t^3+1)) is (t^4+4t)/(2(t^3+1)sqrt(t^3+1))