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

1 answer:

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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0.000309 in scientific notation

trasher [3.6K]

Hey there :)

Writing in scientific notation means a should be
0 ≤ a < 10

Therefore, 0.000309 will be written as
3.09 x 10ⁿ

3 is the 4th decimal place value

so n = - 4

3.04 x 10⁻⁴