What is the derivative of y=47(2t^3-9)^{6/5} ?

The derivative of y=47(2t^3-9)^{6/5} is (1692t^2(2t^3-9)^{1/5})/5

What is the first derivative of y=47(2t^3-9)^{6/5} ?

The first derivative of y=47(2t^3-9)^{6/5} is (1692t^2(2t^3-9)^{1/5})/5

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derivative of

y = 47 (2 t^{3} - 9)^{\frac{6}{5}}

\frac{1692 t ^{2} ( 2 t ^{3} - 9 ) ^{\frac{1}{5}}}{5}

Solution steps

\frac{d}{d t} (47 (2 t^{3} - 9)^{\frac{6}{5}})

Take the constant out:

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

= 47 \frac{d}{d t} ((2 t^{3} - 9)^{\frac{6}{5}})

Apply the chain rule:

\frac{6 ( 2 t ^{3} - 9 ) ^{\frac{1}{5}}}{5} \frac{d}{d t} (2 t^{3} - 9)

= \frac{6 ( 2 t ^{3} - 9 ) ^{\frac{1}{5}}}{5} \frac{d}{d t} (2 t^{3} - 9)

\frac{d}{d t} (2 t^{3} - 9) = 6 t^{2}

= 47 \cdot \frac{6 ( 2 t ^{3} - 9 ) ^{\frac{1}{5}}}{5} \cdot 6 t^{2}

Simplify

47 \cdot \frac{6 ( 2 t ^{3} - 9 ) ^{\frac{1}{5}}}{5} \cdot 6 t^{2} : \frac{1692 t ^{2} ( 2 t ^{3} - 9 ) ^{\frac{1}{5}}}{5}

= \frac{1692 t ^{2} ( 2 t ^{3} - 9 ) ^{\frac{1}{5}}}{5}

\int \frac{3}{5 y - 4} d y

x \to 0 lim (\frac{6 x n - 5 x n - 1 - 7 x n - 2}{- 6 x n - 4 x n - 2})

f (x) = (x^{4} + 3 x^{2} - 2)^{5}

ln (s + 1)

\int \frac{3 - x}{x - 4} d x

What is the derivative of y=47(2t^3-9)^{6/5} ?
The derivative of y=47(2t^3-9)^{6/5} is (1692t^2(2t^3-9)^{1/5})/5
What is the first derivative of y=47(2t^3-9)^{6/5} ?
The first derivative of y=47(2t^3-9)^{6/5} is (1692t^2(2t^3-9)^{1/5})/5