I noticed i did #2 wrong...
f'(x) = 3x^2 - 12x + 5
dy = (3(-2)^2 -12(-2) + 5)(.1)
dy = 4.1
Not sure I am working these problems correctly. I watched some youtube videos and think I figured it out but would like you to verify my work.
Find the differential of the function. Calculate dy for the given valuse of x and dx.
1) y=sqrt(1+x^2), x=0 dx=0.01
f'(x)=1/2(x^2+1)^(-1/2) (2x) = x/sqrt(x^2+1)
dy=f'(x)dx
dy=x/sqrt(x^2+1)dx
dy=0/sqrt(0^2+1)(0.01)=0
2) y=x^3-6x^2+5x-7, x=-2 dx=0.1
f'(x)=1/4 x^4 - 2x^3 + 5/2 x^2 - 7x
dy = f'(x)dx
dy = (1/4(-2)^2-2(-2)^3 +5/2(-2)^2-7(-2))(0.1)
dy = 4.4
3) cos^2(pit), t=1/3 dt=0.05
f'(t) = -2pi cos(pit)sin(pit)
dy=f'(t)dt
dy = -2pi cos(pi/3)sin(pi/3)(0.05)
dy = -2pi (1/2)(sqrt(3)/2)(0.05)
dy = 1/20(sqrt(3)/4)pi
dy = pi sqrt(3)/80 = 0.068
Thank you for looking,
Rod