What is the derivative of 0.01(0.2t^2+4t+64)^{2/3} ?

The derivative of 0.01(0.2t^2+4t+64)^{2/3} is (0.00666…(0.4t+4))/((0.2t^2+4t+64)^{1/3)}

What is the first derivative of 0.01(0.2t^2+4t+64)^{2/3} ?

The first derivative of 0.01(0.2t^2+4t+64)^{2/3} is (0.00666…(0.4t+4))/((0.2t^2+4t+64)^{1/3)}

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derivative of

0.01 (0.2 t^{2} + 4 t + 64)^{\frac{2}{3}}

\frac{0.00666 \dots ( 0.4 t + 4 )}{( 0.2 t ^{2} + 4 t + 64 ) ^{\frac{1}{3}}}

Solution steps

\frac{d}{d t} (0.01 (0.2 t^{2} + 4 t + 64)^{\frac{2}{3}})

Take the constant out:

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

= 0.01 \frac{d}{d t} ((0.2 t^{2} + 4 t + 64)^{\frac{2}{3}})

Apply the chain rule:

\frac{2}{3 ( 0.2 t ^{2} + 4 t + 64 ) ^{\frac{1}{3}}} \frac{d}{d t} (0.2 t^{2} + 4 t + 64)

= \frac{2}{3 ( 0.2 t ^{2} + 4 t + 64 ) ^{\frac{1}{3}}} \frac{d}{d t} (0.2 t^{2} + 4 t + 64)

\frac{d}{d t} (0.2 t^{2} + 4 t + 64) = 0.4 t + 4

= 0.01 \cdot \frac{2}{3 ( 0.2 t ^{2} + 4 t + 64 ) ^{\frac{1}{3}}} (0.4 t + 4)

Simplify

0.01 \cdot \frac{2}{3 ( 0.2 t ^{2} + 4 t + 64 ) ^{\frac{1}{3}}} (0.4 t + 4) : \frac{0.00666 \dots ( 0.4 t + 4 )}{( 0.2 t ^{2} + 4 t + 64 ) ^{\frac{1}{3}}}

= \frac{0.00666 \dots ( 0.4 t + 4 )}{( 0.2 t ^{2} + 4 t + 64 ) ^{\frac{1}{3}}}

x \to 2 lim (3 x + 1)

\int \frac{x - 1}{2 x - x + 1} d x

\frac{d}{d x} (2 x e^{- 2 x})

n = 5 \sum \infty \frac{5 ^{n}}{6 ^{n}}

\frac{d}{d x} (\frac{5}{4 x ^{3} + 4})

What is the derivative of 0.01(0.2t^2+4t+64)^{2/3} ?
The derivative of 0.01(0.2t^2+4t+64)^{2/3} is (0.00666…(0.4t+4))/((0.2t^2+4t+64)^{1/3)}
What is the first derivative of 0.01(0.2t^2+4t+64)^{2/3} ?
The first derivative of 0.01(0.2t^2+4t+64)^{2/3} is (0.00666…(0.4t+4))/((0.2t^2+4t+64)^{1/3)}