Find f'(0.3) for f(x)= the integral from 0 to x of arccos(t)dt
1 answer:
![\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](https://tex.z-dn.net/?f=%5Cbf%20%5Ctextit%7Busing%20the%202nd%20fundamental%20theorem%20of%20calculus%7D%5C%5C%5C%5C%0A%5Ccfrac%7Bdy%7D%7Bdx%7D%5Cdisplaystyle%20%5Cleft%5B%20%5Cint%5Climits_%7B0%7D%5E%7Bx%7D%5C%20cos%5E%7B-1%7D%28t%29dt%20%5Cright%5D%5Cimplies%20cos%5E%7B-1%7D%28x%29%0A%5C%5C%5C%5C%5C%5C%0Af%27%280.3%29%5Ciff%20cos%5E%7B-1%7D%280.3%29%5Capprox%201.26610367277949911126)
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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