Question

d arccos x ceos using the property of derivatives of an inverse function. 14. Find 15. Find all relative extrema of the function 2x- 3x2/3

Answer 1

arc (cos x) = cos^-1 (x) [cosine inverse of x] Let y = cos^-1 (x) (because cos [cos^-1 x] = x) therefore cos y = x equation (1) Differentiating w.r.t x, we get d/dx (cos y) = d/dx x - sin y middot (dy/dx) = 1 therefore dy/dx = -1/sin y Now, we know that sin^2 y + cos^2 y = 1 therefore sin y = squareroot 1 - cos^2 y therefore dy/dx = -1/squareroot 1 - cos^2 y therefore dy/dx = -1/squareroot 1 - x^2 [from (1)] therefore d/dx (cos^-1 x) = -1/squareroot 1 - x^2 \r\n f(x) = 2x -3x^2/3 To find extreme, derivate the function w.r.t. x, and equate it to 0. Therefore Differentiating w.r.t. x., f^'(x) = 2 -3 times 2/3 x^2/3 -1 (because d/dx x^n = n x^n -1) Now, f^'(x) = 0 therefore 2 -2x^-1/3 = 0 implies 2x^1/3 -2/x^1/3 = 0 2 = 2 x^-1/3 [f^'(x) function is undefined at x = 0) \r\n