determine an equation of the tangent to the curve at the given point?

Question

Determine an equation of the tangent to the curve at the given point
(2,1)

y^4 + x^2y^3 = 5

Answer 1

It is the first derivative.

Answer 2

The derivative at that point is -1/4, so the equation is:
y = -1/4 x + 1/2

Answer 3

write it as x=sqrt( (5-y^4)/y^3 )

take derivate of that and plug y=1 into it.

the equations then is x=2+derivative*(y-1)

you can transform it so it reads y = bx+a

Answer 4

take the by-product with d/dx... remembering that d/dx ( f(x) ) = f ' (x) dx / dx = f ' (x) and d /dx ( f(y) ) = f ' (y) dy/dx party : d / dx (x^2) = 2x yet d / dx (y^2) = 2y dy/dx first and 0.33 words are skill regulations, in spite of the indisputable fact that the middle is a product rule (u'v + uv') 4x + 5(y + x*dy/dx) + 10y*dy/dx = 0 4x + 5y + 5x*dy/dx + 10y*dy/dx = 0 4x + 5y + (5x + 10y)*dy/dx = 0 plug contained in the point (2,a million) it rather is (x,y) 4(2) + 5(a million) + (5(2) + 10(a million))dy/dx = 0 8 + 5 + 20*dy/dx = 0 dy/dx = -13/20 so we've the slope and the point, use the point slope kind from algebra (y-y1) = m(x-x1) y-a million = -13/20(x-2) y = -13/20 (x-2) + a million... i ought to leave it in simple terms how that's