What is dy/dx when y=ln(6x)?

This is a common question on A2 Maths papers as it tests both the ability to use the chain rule and the ability to differentiate natural logarithms. y = ln(6x) Alarm bells should start ringing whenever a question asks you to differentiate an equation and there is a function inside a function. In this case, the chain rule should be used. Usually here I would say write down the chain rule formula, however often this can make things more confusing. For clarity though, the formula is: dy/dx = (du/dx)*(dy/du) To begin working this out, let: u = 6x     y = ln(u)    <------ rewrite y in terms of u here, usually u is whatever is in the bracket! Next, differentiate both of these: du/dx = 6  dy/du = 1/u  <------ this is the standard differentation of a natural log (I can prove this on request!) Now, use the chain rule formula to find dy/dx, or if you're like me and get more confused using this, then just remember to multiply both the derivatives together that you've just calculated! So: dy/dx = 6/u This is not in terms of x like the original equation is, so we need to replace u with 6x: dy/dx = 6/6x = 1/x