Given y = 3x^(1/2) - 6x + 4, x > 0. 1) Find the integral of y with respect to x, simplifying each term. 2) Differentiate the equation for y with respect to x.
- When integrating remember the general rule: x^n --> x^(n+1)/(n+1). Now look at each term separately. 3x^(1/2) --> 3x^(3/2)/(3/2) = 2x^(3/2)-6x --> -6x^(2)/2 = -3x^24 = 4x^(0) --> 4x^(1)/1 = 4xCombining we find that the integral of y with respect to x equals, 2x^(3/2) - 3x^(2) + 4x + cn.b. Need to remember the constant "+c"
2) When differentiating remember the general rule: x^n --> x^(n-1) * n = nx^(n-1). Again now look at each term separately.3x^(1/2) --> 3x^(-1/2) * (1/2) = (3/2)x^(-1/2)-6x = -6x^(1) --> -6x^(0) * 1 = -6x^(0) = -64 --> 0Combining we find that the differential of y with respect to x equals, (3/2)x^(-1/2) - 6
SO
