Answer 1

dy/dx when 4x^2 + 5x^4y - 4y^2 = 2322

\[4\frac{d}{dx}x^2 + 5\frac{d}{dx}(x^4y)- 4\frac{d}{dx}y^2 = \frac{d}{dx}2322\]

Answer 2

\[8x\frac{dx}{dx} + 5(x^4\frac{dy}{dx}+4x^3y\frac{dx}{dx})- 8y\frac{dy}{dx} = 0\]
\[(5x^4-8y)\frac{dy}{dx}= -(8x+20x^3y)\]
\[\frac{dy}{dx}= \frac{8x+20x^3y}{8y-5x^4}\]
To find the slope at (3,6), substitute x=3, y=6 into the above equation.
Use the point-slope formula to find the equation of the line tangent to the curve at point (3,6)
y-y0= m(x-x0) where m is the slope just found and (x0,y0) is (3,6)