What is the derivative of (t+4)^{(2/3)}(2t^2-4)^3 ?

The derivative of (t+4)^{(2/3)}(2t^2-4)^3 is (2(2t^2-4)^3+36t(2t^2-4)^2(t+4))/(3(t+4)^{1/3)}

What is the first derivative of (t+4)^{(2/3)}(2t^2-4)^3 ?

The first derivative of (t+4)^{(2/3)}(2t^2-4)^3 is (2(2t^2-4)^3+36t(2t^2-4)^2(t+4))/(3(t+4)^{1/3)}

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derivative of

(t + 4)^{(\frac{2}{3})} (2 t^{2} - 4)^{3}

\frac{2 ( 2 t ^{2} - 4 ) ^{3} + 36 t ( 2 t ^{2} - 4 ) ^{2} ( t + 4 )}{3 ( t + 4 ) ^{\frac{1}{3}}}

Solution steps

\frac{d}{d t} ((t + 4)^{\frac{2}{3}} (2 t^{2} - 4)^{3})

Apply the Product Rule:

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

f = (t + 4)^{\frac{2}{3}}, g = (2 t^{2} - 4)^{3}

= \frac{d}{d t} ((t + 4)^{\frac{2}{3}}) (2 t^{2} - 4)^{3} + \frac{d}{d t} ((2 t^{2} - 4)^{3}) (t + 4)^{\frac{2}{3}}

\frac{d}{d t} ((t + 4)^{\frac{2}{3}}) = \frac{2}{3 ( t + 4 ) ^{\frac{1}{3}}}

\frac{d}{d t} ((2 t^{2} - 4)^{3}) = 12 t (2 t^{2} - 4)^{2}

= \frac{2}{3 ( t + 4 ) ^{\frac{1}{3}}} (2 t^{2} - 4)^{3} + 12 t (2 t^{2} - 4)^{2} (t + 4)^{\frac{2}{3}}

Simplify

\frac{2}{3 ( t + 4 ) ^{\frac{1}{3}}} (2 t^{2} - 4)^{3} + 12 t (2 t^{2} - 4)^{2} (t + 4)^{\frac{2}{3}} : \frac{2 ( 2 t ^{2} - 4 ) ^{3} + 36 t ( 2 t ^{2} - 4 ) ^{2} ( t + 4 )}{3 ( t + 4 ) ^{\frac{1}{3}}}

= \frac{2 ( 2 t ^{2} - 4 ) ^{3} + 36 t ( 2 t ^{2} - 4 ) ^{2} ( t + 4 )}{3 ( t + 4 ) ^{\frac{1}{3}}}

x \to 2 lim (x^{2} + 12)

\frac{d ^{2}}{d x ^{2}} (\frac{7}{8 x ^{6}})

f (x) = 6 x - 32 x, at x = 4

f (x) = \frac{x ^{2} + sec ( x )}{x - 2 e ^{x}}

f (x) = \frac{x ^{2} + 3}{x}

What is the derivative of (t+4)^{(2/3)}(2t^2-4)^3 ?
The derivative of (t+4)^{(2/3)}(2t^2-4)^3 is (2(2t^2-4)^3+36t(2t^2-4)^2(t+4))/(3(t+4)^{1/3)}
What is the first derivative of (t+4)^{(2/3)}(2t^2-4)^3 ?
The first derivative of (t+4)^{(2/3)}(2t^2-4)^3 is (2(2t^2-4)^3+36t(2t^2-4)^2(t+4))/(3(t+4)^{1/3)}