What is the derivative of h=(t+1)^{2/3}(2t^2-1)^3 ?

The derivative of h=(t+1)^{2/3}(2t^2-1)^3 is (2(2t^2-1)^3+36t(2t^2-1)^2(t+1))/(3(t+1)^{1/3)}

What is the first derivative of h=(t+1)^{2/3}(2t^2-1)^3 ?

The first derivative of h=(t+1)^{2/3}(2t^2-1)^3 is (2(2t^2-1)^3+36t(2t^2-1)^2(t+1))/(3(t+1)^{1/3)}

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derivative of

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

\frac{2 ( 2 t ^{2} - 1 ) ^{3} + 36 t ( 2 t ^{2} - 1 ) ^{2} ( t + 1 )}{3 ( t + 1 ) ^{\frac{1}{3}}}

Solution steps

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

Apply the Product Rule:

(f \cdot g)^{'} = f^{'} \cdot g + f \cdot g^{'}

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

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

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

\frac{d}{d t} ((2 t^{2} - 1)^{3}) = 12 t (2 t^{2} - 1)^{2}

= \frac{2}{3 ( t + 1 ) ^{\frac{1}{3}}} (2 t^{2} - 1)^{3} + 12 t (2 t^{2} - 1)^{2} (t + 1)^{\frac{2}{3}}

Simplify

\frac{2}{3 ( t + 1 ) ^{\frac{1}{3}}} (2 t^{2} - 1)^{3} + 12 t (2 t^{2} - 1)^{2} (t + 1)^{\frac{2}{3}} : \frac{2 ( 2 t ^{2} - 1 ) ^{3} + 36 t ( 2 t ^{2} - 1 ) ^{2} ( t + 1 )}{3 ( t + 1 ) ^{\frac{1}{3}}}

= \frac{2 ( 2 t ^{2} - 1 ) ^{3} + 36 t ( 2 t ^{2} - 1 ) ^{2} ( t + 1 )}{3 ( t + 1 ) ^{\frac{1}{3}}}

\int \frac{x x}{2} d x

f (x) = \frac{( 4 x + x ^{2} )}{( 2 + x ) ^{2}}

\int \frac{14 - 14 x}{1 - x} d x

\frac{d y}{d t} = - \frac{t}{e ^{y + t^{2}}}

\int \frac{3 + x + x}{x} d x

What is the derivative of h=(t+1)^{2/3}(2t^2-1)^3 ?
The derivative of h=(t+1)^{2/3}(2t^2-1)^3 is (2(2t^2-1)^3+36t(2t^2-1)^2(t+1))/(3(t+1)^{1/3)}
What is the first derivative of h=(t+1)^{2/3}(2t^2-1)^3 ?
The first derivative of h=(t+1)^{2/3}(2t^2-1)^3 is (2(2t^2-1)^3+36t(2t^2-1)^2(t+1))/(3(t+1)^{1/3)}