Find the equation of the tangent to the curve y=ln(3x-2) + 4y=ln(3x−2)+4 at the point where x=1x=1?
Thanks!
Thanks!
1 Answer
Nov 27, 2017
Explanation:
•color(white)(x)m_(color(red)"tangent")=dy/dx" at x = a"∙xmtangent=dydx at x = a
"differentiate using the "color(blue)"chain rule"differentiate using the chain rule
"given "y=f(g(x))" then"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]}