4)If f(2)=3 and f'(2)=-1, find equation of the tangent line when x=2.(In point slope form)
5)Find dy/dx, x^4+4x^2y^3+y^2=2y then evaluate the derivative at the point (1,-1).
6)Find derivative of y=2
y-3 = 1*(x-2)
y = x-2+3 or
dy/dx +2y dy/dx = 2 dy/dx
dy/dx = -4x(x2+2y3)/(12x2y2+2y-2)
at (1,-1) this becomes
= -4(1-2)/(12-2-2) =4/8 = 1/2
at x=2, dy/dx = =-8*(4+2y3)/(48y2+2y-2) where (2,y) lies on the given curve. The determination of y is by solving
a cubic equation in y namely 16+16y3+y2-2y=0
Content will be erased after question is completed.
Enter the email address associated with your account, and we will email you a link to reset your password.
Forgot your password?