What is the d/(dt)((t+1)^{2/3}(2t^2-1)^3) ?

The d/(dt)((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 d/(dt)((t+1)^{2/3}(2t^2-1)^3) ?

The first d/(dt)((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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\frac{d}{d 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 + 6}{x ^{2} + 11 x + 18}) d x

\frac{d}{d x} (x^{6} + 3 x^{4} + 7 x^{2})

cos (π t)

f^{^{''}} (x) = x^{\frac{- 3}{2}}, f^{^{'}} (4) = 2, f (0) = 0

\frac{\partial}{\partial x} (cos (x^{6} - 7 y))

What is the d/(dt)((t+1)^{2/3}(2t^2-1)^3) ?
The d/(dt)((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 d/(dt)((t+1)^{2/3}(2t^2-1)^3) ?
The first d/(dt)((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)}