f'(x) will be found using the Power Rule. The Power Rule states that given a term Ax^n, the derivative of that term with respect to x is nAx^(n-1). It may seem difficult to use that here; however, it becomes easier when we recall that sqrtx = x^(1/2). We must also recall that if our term is simply Cx, there is an exponent on the x; specifically, 1. Any number to the first power is simply that number. Thus Cx = Cx^1, and consequently the derivative with respect to x is 1*Cx^0 = 1*C*1 = C. With that in mind...
f(x) = 7x - 32sqrtx = 7x -32x^(1/2) -> f'(x) = 7 - (1/2)32x^(-1/2) = 7 - 16/x^(1/2) = 7 - 16/sqrt(x)
f'(x) is the equation for the slope of the line tangent to our original curve at any point. In order to find the equation for the line at a given point a, we must first know f(a).
f(a) = f(4) = 7(4)-32sqrt(4) = 28 -32*2 = 28-64 = -36
Thus, our original curve passes through the point (4, -36). We can now use point-slope form to find the equation for the line tangent to our curve at x=4.
y-y_1 = m(x-x_1) -> y+36 = (7-16/sqrtx)(x-4) -> y = (7-16/sqrtx)(x-4) -36
The equation for the line tangent to the curve f(x) = 7x - 32sqrtx is y = (7-16/sqrtx)(x-4) -36
