Question

What is the equation of the tangent line of `f(x)=(x-2)/(x-3)+lnx/(3x)-x` at `x=2`?

Answer 1

`y=-1/12(23x+xln2)+(2ln2+11)/6`

Explanation:

To find the equation of the tangent line of the function we can start by taking its derivative, which will give us the slope of the tangent line to the curve.

`f(x)=(x-2)/(x-3)+(lnx)/(3x)-x`

I prefer to rewrite using products rather than quotients to avoid the quotient rule, but either approach will work.

`=>f(x)=(x-2)(x-3)^-1+(lnx)(3x)^-1-x`

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

`=(2-x)/(x-3)^2+1/(x-3)+1/(3x^2)-(3lnx)/(9x^2)-1`

You could choose to simplify, but since we're given a value I will save time by plugging it in now and evaluating.

`f'(2)=0-1+1/12-(3ln2)/36-1`

`=-(23+ln2)/12`

This is `m` in the point-slope form of a linear equation, `y-y_1=m(x-x_1)`.

We can find `(x_1,y_1)`, a point which lies on the tangent line. This point also lies on the path of the function, so at `x=2`, we have `f(2)=y=(ln2)/6-2`.

We can put these values into the generic equation above.

`y-((ln2)/6-2)=-(23+ln2)/12(x-2)`

`=>y=-(23+ln2)/12(x-2)+(ln2)/6-2`

`y=-((23+ln2)(x-2))/12+ln2/6-2`

`y=((-23-ln2)(x-2))/12+ln2/6-2`

`y=(-23x+46+2ln2-xln2)/12+ln2/6-2`

`y=(-23x+46+2ln2-xln2+2ln2-24)/12`

`y=(-23x+22+4ln2-xln2)/12`

`y=-1/12(23x+xln2)+(2ln2+11)/6`