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Find equation of the tangent line to the curve at the given point y = ln(x) at x0 = 8

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The gradient of a tangent line to the curve is give by the first derivative.

`y = lnx`

`y' = 1/x`


So the gradient of any tangent line to curve `y = lnx` is given by `1/x` .


Gradient at x = 8 of the tangent = 1/8


The tangent equation is `y = mx+c`

Now we have found gradient (m) = 1/8

So `y = 1/8x+c`


The tangent line and curve will have same y coordinates at x = 8

For the curve

`y = lnx`

at x = 8

`y = ln8`


For the tangent line

`y = 1/8x+c`

at x = 8

`ln8 = 1/8*8+c`

 `c = ln8-1`


So the equation of the tangent is;

`y = 1/8x+ln8-1`




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