Assume that hx fx3 where f is a differentiable function if

Assume that h(x) = (f(x))^3, where f is a differentiable function. if f(0) = -1/2 and f\'(0) = 8/3 determine an equation of the line tangent to the graph of h at x = 0. Consider a function y = h(x) defined implicitly by x^2 + xy + y^3 = 1 compute h(1) h\'(1) h\'\'(1) h\'\'\'(1)

Solution

(1) h(x)= (f(x))³

so h\'(x)= 3 (f(x))² f\'(x) by chain rule

so h\'(0) = 3 x 1/4 x 8/3= 2

h(0) = -1/8

The slope of the tangent =h\'(0) =2 and it passes through (0,-1/8).

So the required equation is

(y+1/8) = 2(x-0)

or 8y+1=2x

(2) x² +xy+y³ =1........................(1)

y=h(x) defined implicitly by (1)

Putting x=1, y+y³ =0 implying y =0 (as the other roots are not real)

So h(1)=0

Differentiate (1) wrt to x to get

2x+ xy\'+y+ 3y² y\'=0......................(2)

Put x=1 and y=0 ,to get

                   h\'(0) = -2

Differentiating (2),

2+ y\' +xy\'\'+y\'+6y(y\')² +3y² y\'\'=0........................(3)

Putting x=1,y=0, y\'=-2, yields

y\'\'(0)= h\'\'(1)=-2

Differentiating (3), we get

2y\'\'+xy\'\'\'+y\'\'+6(y\')³ +12yy\'y\'\'+6yy\'y\'\' +3y² y\'\'\'=0

Substituting x=1, y=0, y\'=-2 and y\'\'=-2

-4+y\'\'\'-2-48=0

so y\'\'\'(1)=h\'\'\'(1) = 54