#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#
