Question

How do you use implicit differentiation to find an equation of the tangent line to the curve `x^2 + 2xy − y^2 + x = 39` at the given point (5, 9)?

Answer 1

`29x-8y=73`

Explanation:

Differentiate the equation with respect to `x`:

`d/dx (x^2+2xy-y^2+x) = 0`

`2x +2y+2xdy/dx -2ydy/dx +1 = 0`

`(2x-2y) dy/dx = -1-2x-2y`

`dy/dx = (1+2x+2y)/(2y-2x)`

The equation of the tangent line is:

`y= y_0 + y'(x_0)(x-x_0)`

where `x_0 = 5`, `y_0 = 9` and:

`y'(x_0) = (1+10+18)/(18-10) = 29/8`

then:

`y = 9+29/8(x-5)`

`y=29/8x -73/8`

or:

`29x-8y=73`

Answer 2

`29x-8y-73=0`
is the equation of the tangent

Explanation:

Given:
`x^2+2xy-y^2+x=39`
when x=5, and y=9
`x^2+2xy-y^2+x=5^2+2xx5xx9-9^2+5`
`=25+90-81+5=(25+5)+(90-81)`
`=30+9=39`
`x^2+2xy-y^2+x=39`
verified

We have
`x^2+2xy-y^2+x=39`
Differentiating both sides wrt x
`2x+2(xdy/dx+y)-2ydy/dx+1=0`
Substituting x=5 and y=9
`2xx5+2(5xxdy/dx+9)-2xx9dy/dx+1=0`

`10+10dy/dx+18-18dy/dx+1=0`
`(10+18+1)+(10-18)dy/dx=0`
`(28+1)-8dy/dx=0`

`29=8dy/dx`

`dy/dx=29/8`
Slope of the tangent is
`m=29/8`
Passing through the point
`P-=(5,9)`
Equation of the tangent is
`(y-9)/(x-5)=29/8`

`8(y-9)=29(x-5)`
`8y-72=29x-145`
`27x-8y+72-145=0`

`29x-8y-73=0`