find an equation of the tangent line to the graph of the fun

Solution

First let us simplify the equation given for f(x) f(x) = x^2*e^{(3/2)ln(x)} = x^2*e^[ln{x^(3/2)}] ........ [Since a*ln(x) = ln{x^(a)}] = x^2*x^(3/2) ........ [Since e^{ln(x)} = x] = x^(7/2) Therefore, d(f(x))/dx = (7/2)*x^(7/2-1) = (7/2)*x^(5/2) At x = 1, the value of f(x) = 1^(7/2) = 1 and the value of d(f(x))/dx = (7/2)*1^(5/2) = 7/2 Therefore the slope of the tangent is 7/2 and it passes through the point (1,1) The equation of a line with slope m and passing through the point (x1,y1) is given by: (y-y1) = m*(x-x1) Therefore the equation of the tangent is: (y-1) = 7/2*(x-1) 2y-2 = 7x -7 7x-2y = 5