# 14a) find the slope of the tangent to the curve a...

14a) find the slope of the tangent to the curve at the pont where .
14b. find the equations of the tangent lines at the pont (1,1)and (4, 1/2)

1. (a) then so the slope at the point where x = a is (B)
solpe at the point (1,1) is = -1/2
so the equation of the tangent line is (y-1) =(-1/2)(x-1)
y = -x/2 +3/2
slope at the point (4,1/2) is = -1/16
then the equation of the tangent line at ( 4,1/2) is

(y-1/2)=(-1/16)(x-4)

y = -x/16+3/4
2. Quotient rule: (f/g)'=(gf'-fg')/g^2
The slope of the tangent line is the derivative of thefunction, so
f'(x)=y'=sqrt(x)*1'-1sqrt(x)'/sqrt(x)^2=-sqrt(x)/sqrt(x)^2,and install a for x, which doesn't mean anything until you get topart b, where you install the points (x,y) for their respectivevalues in the function.
3. 14 b = x^(-1/2)
y' = -1/2*[x^(-1/3)......................1
Put x= 1 in eqn 1,
y ' = slope = -1/2
Equation of tangent => (y-y1)=m(x-x1)
Here m = -1/2 and x1,y1 = 1,1 Substitutethe valus and get the answers. Similarly you can do the otherone