Question

Find the slope of the line tangent to the graph of (x^2)-((xy)^(1/2))=(2y^2)+3x at the point (4,1).

Answer 1

Compute
`(\mathrm dy)/(\mathrm dx)` using implicit differentiation:

`x^2-(xy)^(1/2)=2y^2+3x`

`(\mathrm d(x^2))/(\mathrm dx)-(\mathrm d(xy)^(1/2))/(\mathrm dx)=(\mathrm d(2y^2))/(\mathrm dx)+(\mathrm d(3x))/(\mathrm dx)`

`2x-\frac1{2(xy)^(1/2)}\left((\mathrm dx)/(\mathrm dx)y+x(\mathrm dy)/(\mathrm dx)\right)=4y(\mathrm dy)/(\mathrm dx)+3`

`2x-\frac1{2(xy)^(1/2)}\left(y+x(\mathrm dy)/(\mathrm dx)\right)=4y(\mathrm dy)/(\mathrm dx)+3`

`2x-\frac12\left(\frac yx\right)^(1/2)-\frac12\left(\frac xy\right)^(1/2)(\mathrm dy)/(\mathrm dx)=4y(\mathrm dy)/(\mathrm dx)+3`

`2x-3-\frac12\left(\frac yx\right)^(1/2)=\left(4y+\frac12\left(\frac xy\right)^(1/2)\right)\right)(\mathrm dy)/(\mathrm dx)`

`4x-6-\left(\frac yx\right)^(1/2)=\left(8y+\left(\frac xy\right)^(1/2)\right)\right)(\mathrm dy)/(\mathrm dx)`

`(\mathrm dy)/(\mathrm dx)=(4x-6-\left(\frac yx\right)^(1/2))/(8y+\left(\frac xy\right)^(1/2))`

`(\mathrm dy)/(\mathrm dx)=(4x(xy)^(1/2)-6(xy)^(1/2)-y)/(8y(xy)^(1/2)+x)`

Plug in x = 4 and y = 1 to find that the tangent line's slope is 19/20.