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A curve C has equation y = x^2 − 2x − 24 x^(1/2), x > 0. Find dy/dx and d^2y/dx^2. Verify that C has a stationary point when x = 4

Using the differentiation rule that d (Ax^b)/dx = Abx^(b-1) we find dy/dx = 2x -2 -12x^(-1/2).Similarly, taking care to see that the -2 term becomes zero since it is not dependent on x, we haved^2y/dx^2 = 2 + 6x^(-3/2).By substituting the value x = 4 into our expression of dy/dx we have2x4 -2 -12x(4^(-1/2)) = 0. Hence we have a stationary point at the value x = 4.

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A curve C has equation y = x^2 − 2x − 24 x^(1/2), x > 0. Find dy/dx and d^2y/dx^2. Verify that C has a stationary point when x = 4

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