Question

What is the equation of the tangent to the line `sf(y=ln(x^2-8)` at the point (3,0) ?

Answer 1

`y=6x-18`

Explanation:

The equation of the tangent is `y=6x`- 18.

To find the equation of the tangent we must find the first derivative of the function. To do this we must use the chain rule (function of a function).

This states that:

`(dy)/dx=(dy)/(dt)*(dt)/(dx)`

The outer layer of the chain is the `ln` bit. The inner layer is the `(x^2-8)` bit.

Let `t=(x^2-8)`

So `y=lnt`

So `(dy)/(dt)=(1)/(x^2-8)`

(This is because the derivative of `lnx = 1/x`)

And `(dt)/(dx)=2x`

So the product of the 2 derivatives `(dy)/(dx)` becomes-->
`(dy)/(dx)=(1)/(x^2-8)*2x`

This gives the gradient of the line at a particular value of `x`. At (3,0) `x=3` so the gradient is :

`(2*3)/(3^3-8)=6`

The general equation for a straight line is `y=mx+c` where `m` is the gradient and `c` is the intercept.

To get `c` we can plug in the values we now know:

`0=6*3+c`

From which `c` = -18

So the equation of the tangent is:

`y=6x-18`

The situation looks like this:

graph{(ln(x^2-8)-y)(6x-18-y)=0 [-10, 10, -5, 5]}