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

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

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To find the equation of the tangent to the curve y=1/(3x) at the point where x=1/6, we need to find the slope of the tangent line at that point.

First, we find the derivative of the function y=1/(3x) using the power rule for differentiation.

The derivative of y with respect to x is given by dy/dx = -1/(3x^2).

Next, we substitute x=1/6 into the derivative to find the slope of the tangent line at that point.

dy/dx = -1/(3(1/6)^2) = -1/(3(1/36)) = -1/(1/12) = -12.

Therefore, the slope of the tangent line at x=1/6 is -12.

Using the point-slope form of a linear equation, y - y1 = m(x - x1), where (x1, y1) is the point on the curve, we substitute x=1/6, y=1/(3(1/6)) = 2 into the equation.

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

Simplifying the equation, we get y - 2 = -12x + 2.

Rearranging the equation, we find the equation of the tangent line to be y = -12x + 4.

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