Question

Find the equation of the tangent to the curve `y=ln(3x-2) + 4` at the point where `x=1`?

Thanks!

Answer 1

`y=3x+1`

Explanation:

`•color(white)(x)m_(color(red)"tangent")=dy/dx" at x = a"`

`"differentiate using the "color(blue)"chain rule"`

`"given "y=f(g(x))" then"`

`dy/dx=f'(g(x))xxg'(x)larr"chain rule"`

`y=ln(3x-2)+4`

`rArrdy/dx=1/(3x-2)xxd/dx(3x-2)=3/(3x-2)`

`dy/dx(x=1)=3/1=3`

`y=ln(3-2)+4=0+4=4rArr(1,4)`

`rArry-4=3(x-1)larr"point-slope form"`

`rArry=3x+1larr"slope-intercept form"`
graph{(y-ln(3x-2)-4)(y-3x-1)=0 [-20, 20, -10, 10]}