Question

Find the equation of the tangent line to the curve (piriform)
y^2=x^3(4−x)
at the point (2,16−−ã).

a. Find dy/dx at x=2.
dy/dx=

b. Write the equation of the tangent line to the curve.

Answer 1

y^2 = 4x^3 - x^4
2yy' = 12x^2 - 4x^3
y' = (6x^2 - 2x^3)/y

y(2) = 4
y'(2) = (6*4-2*8)/4 = 2

So, now you have a point and a slope. The tangent line at (2,4) is

y-4 = 2(x-2)

Answer 2

2y dy/dx = x^3 (-1) + 3x^2 (4-x)

dy/dx = (-x^3 + 12x^2 - 3x^3)/(2y)
= (6x^2 - 2x^3)/y

I can't make out your point (2, 16−−ã)
but it should be easy for you
just plug in x=2 and y = whatever into the dy/dx
and that becomes your slope

then use the grade 9 way you learned to find the equation of a line with a given slope and a given point.