1 year ago

30 pts+

Mathematics

1 answer:

Veseljchak [2.6K]1 year ago

7 0

Answer:

I dont know the answer I'm just scrolling down this app because I like going through my brothers phone exceptionally mommy and daddy wrestling

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Find f'(0.3) for f(x)= the integral from 0 to x of arccos(t)dt

Lady_Fox [76]

$\bf \textit{using the 2nd fundamental theorem of calculus}\\\\
\cfrac{dy}{dx}\displaystyle \left[ \int\limits_{0}^{x}\ cos^{-1}(t)dt \right]\implies cos^{-1}(x)
\\\\\\
f'(0.3)\iff cos^{-1}(0.3)\approx 1.26610367277949911126$

now.. 0.3 is just a value...we'e assuming Radians for the inverse cosine, so, if you check, make sure your calculator is in Radian mode

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2 years ago

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Answer:

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Step-by-step explanation:

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