How do you find the equation of the tangent to the curve #y=1/(3x)# at the point where #x=1/6#?

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Answer 1 · 2016-12-11T16:41:55

#y-2=-12(x-1/6)#

#y=-12x+4#

To find the equation, you need the point that the line passes through, and the slope of the line.

Point:
#x=1/6#
#y(1/6)=1/[3(1/6)]=2#
#(1/6,2)#

Slope:
#y(x)=1/(3x)#
#y(x)=1/3x^(-1)#

#y'(x)=-1/3x^(-2)#

#y'(x)=frac{-1}{3x^2}#

#f'(1/6)=frac{-1}{3(1/6)^2}#
#=-1/(1/12)#
#=-12#

Equation (point-slope form):
#y-2=-12(x-1/6)#

Equation (slope-intercept from):
#y=-12x+2+2#

#y=-12x+4#