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

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1 Answer
Jim G.
Nov 27, 2017

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]}

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