Question

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

Answer 1

`\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