Hi can someone please help me with this question:
find the equation of the tangent to the curve y=1/3x at the point where x=1/6.
Thanks 
It's a very similar process to Jess'
To find the equation of the tangent (which you know will be a line) you need to know the gradient and a point on the line. From there you can use y-y1 = m(x-x1)
To find the gradient, simply find the derivative and substitute in the x value (in this case, 1/6)
To find a point on the line, you know the tangent at x=1/6 touches the curve at x=1/6, so substitute this x value into the equation y=1/3x to find the y value where x=1/6. This will be a point on the line.
Hey there, i double checked the question but there arent any powers anywhere. But do you think it will work if (1/3x) is differentiated using the quotient rule?
I think it was misinterpreted as (1/3)x which is a line, instead of 1/(3x) which is a curve.
You shouldn't need to use the quotient rule - use standard derivatives instead.
1/(3x) can be rewritten as

The quotient rule is best when there are two different expressions involving x on the top and bottom of a fraction, and product rule is for when two expressions involving x are multiplied together.
Note: the quotient and product rule will work even when it's not necessary to use them, it's just time consuming.