Question #0c7df

1 Answer
Sep 26, 2017

g'(pi/3)=5/2-7/2sqrt3
h'(pi/3)=(7-5sqrt(3))/50

Explanation:

f(pi/3)=5 and f'(pi/3)=-7

g(x)=f(x)sinx
differentiate with respect to x
g'(x)=sinxd/dx(f(x))+f(x)d/dx(sinx)
g'(x)=sinx.f'(x)+f(x)cosx

then
g'(pi/3)=sin(pi/3)f'(pi/3)+f(pi/3)cos(pi/3)
Put all values of f(pi/3), f'(pi/3) and sin(pi/3) and cos(pi/3)
g'(pi/3)=sqrt3/2.(-7)+(5)(1/2)=5/2-7/2sqrt3

Now
function h(x)=cosx/(f(x))
differentiate with respect to x
h'(x)=(f(x).d/dx(cosx)-cosx.d/dx(f(x)))/(f(x))^2
h'(x)=(f(x).(-sinx)-cosx.(f'(x)))/(f(x))^2
h'(x)=(-f(x).(sinx)-cosx.(f'(x)))/(f(x))^2

h'(pi/3)=(-f(pi/3).(sin(pi/3))-cos(pi/3).(f'(pi/3)))/(f(pi/3))^2

h'(pi/3)=(-(5).(sqrt3/2)-(1/2).(-7))/(5)^2
h'(pi/3)=(-5/2sqrt3+7/2)/25=(7-5sqrt(3))/50