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