Answer these

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

4.

y-3 = 1*(x-2)

y = x-2+3 or

y =x+1

5.

x^{4}+4x^{2}y^{3}+y^{2}=2y

4x^{3}+8xy^{3}+4x^{2}*3y^{2} dy/dx +2y dy/dx = 2 dy/dx

dy/dx *(12x^{2}y^{2}+2y-2) = -(4x^{3}+8xy^{3})

dy/dx = -4x(x^{2}+2y^{3})/(12x^{2}y^{2}+2y-2)

at (1,-1) this becomes = -4(1-2)/(12-2-2) =4/8 = 1/2

at x=2, dy/dx = =-8*(4+2y^{3})/(48y^{2}+2y-2) where (2,y) lies on the given curve. The determination of y is by solving

a cubic equation in y namely 16+16y^{3}+y^{2}-2y=0

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