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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