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

1 answer:

8 0

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

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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