What is the (((t^2+1)^{3/2})/3)^' ?

The (((t^2+1)^{3/2})/3)^' is t(t^2+1)^{1/2}

What is the first (((t^2+1)^{3/2})/3)^' ?

The first (((t^2+1)^{3/2})/3)^' is t(t^2+1)^{1/2}

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\frac{( t ^{2} + 1 ) ^{\frac{3}{2}}}{3}^{^{'}}

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

Solution steps

\frac{( t ^{2} + 1 ) ^{\frac{3}{2}}}{3}^{'}

Take the constant out:

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

= \frac{1}{3} ((t^{2} + 1)^{\frac{3}{2}})^{^{'}}

Apply the chain rule:

\frac{3 ( t ^{2} + 1 ) ^{\frac{1}{2}}}{2} (t^{2} + 1)^{'}

= \frac{3 ( t ^{2} + 1 ) ^{\frac{1}{2}}}{2} (t^{2} + 1)^{'}

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

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

Simplify

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

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

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What is the (((t^2+1)^{3/2})/3)^' ?
The (((t^2+1)^{3/2})/3)^' is t(t^2+1)^{1/2}
What is the first (((t^2+1)^{3/2})/3)^' ?
The first (((t^2+1)^{3/2})/3)^' is t(t^2+1)^{1/2}