Question

Suppose the curve defined by the equation 3x^2 = xy^2 + 1. Answer the following.

a) Calculate dy/dx . Explain its meaning.
b) Find the tangent line to the curve at the point (1, 2)

please help me!

Answer 1

`(\mathrm d)/(\mathrm dx)\left[3x^2\right]=(\mathrm d)/(\mathrm dx)\left[xy^2+1\right]`

`\implies 6x=y^2+2xy(\mathrm dy)/(\mathrm dx)`

`\implies(\mathrm dy)/(\mathrm dx)=(6x-y^2)/(2xy)`

The tangent line to the curve at
`(1,2)` has slope
`(6*1-2^2)/(2*1*2)=\frac12`. In point-slope form, the line's equation is


`y-2=\frac12(x-1)\implies y=\frac12x+\frac32`