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Iteru [2.4K]
4 years ago
13

What is the derivative of the 5th root of t^3?

Mathematics
1 answer:
mariarad [96]4 years ago
6 0

\bf y=t^{\frac{3}{5}}\implies \cfrac{dy}{dt}=\stackrel{\textit{using the power rule}}{\cfrac{3}{5}t^{\frac{3}{5}-1}}\implies \cfrac{dy}{dt}=\cfrac{3}{5}t^{-\frac{2}{5}}
\\\\\\
\cfrac{dy}{dt}=\cfrac{3}{5t^{\frac{2}{5}}}\implies \cfrac{dy}{dt}=\cfrac{3}{5\sqrt[5]{t^2}}

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8 x 2/3 x (-6) x (3/4) x (-1)
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Simplify the following:

(-6 (-8)×2×3 x x x x)/(3×4)

(8 x×2 x (-6) x×3 x (-1))/(3×4) = (8 x^4×2 (-6)×3 (-1))/(3×4):

(-6 (-8)×2×3 x^4)/(3×4)

8×2 = 16:

(-6 (-3)16 x^4)/(3×4)

16 (-6) = -96:

(-3-96 x^4)/(3×4)

-96 (-3) = 288:

(288 x^4)/(3×4)

3×4 = 12:

(288 x^4)/12

|  

1 | 2 | | 2 | 4

| 2 | 8 | 8

- | 2 | 4 |  

| | 4 | 8

| - | 4 | 8

| | | 0:

Answer:  24 x^4


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