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

The d/(dt)((t+3)^{2/3}(2t^2-3)^3) is (2(2t^2-3)^3+36t(2t^2-3)^2(t+3))/(3(t+3)^{1/3)}

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

The first d/(dt)((t+3)^{2/3}(2t^2-3)^3) is (2(2t^2-3)^3+36t(2t^2-3)^2(t+3))/(3(t+3)^{1/3)}

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

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

Solution steps

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

Apply the Product Rule:

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

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

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

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

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

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

Simplify

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

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

f (t) = 5 t - 9 t^{2}

\frac{\partial}{\partial x} (x^{2} - x y + y^{2} + 3 x - 2 y + 1)

y^{2} = 2 x y y^{^{'}}

x \to - 1 lim (2)

\frac{\partial}{\partial x} (\frac{x}{x ^{2}})

What is the d/(dt)((t+3)^{2/3}(2t^2-3)^3) ?
The d/(dt)((t+3)^{2/3}(2t^2-3)^3) is (2(2t^2-3)^3+36t(2t^2-3)^2(t+3))/(3(t+3)^{1/3)}
What is the first d/(dt)((t+3)^{2/3}(2t^2-3)^3) ?
The first d/(dt)((t+3)^{2/3}(2t^2-3)^3) is (2(2t^2-3)^3+36t(2t^2-3)^2(t+3))/(3(t+3)^{1/3)}