How do you differentiate y=ln(x)

I would use the fact that ln is the inverse function of the exponential function e^x to re-write the equation as x=e^y. This can be directly seen by just putting e^y=e^(lnx). Since the definition of a ln(a)=b is that e^b=a it means that e^(ln(a))=a.Now we can implicitly differentiate our expression. The best way to think about this is to think about applying the same operator d/dx to both sides where this simply means differentiate wrt x. The LHS becomes e^y.dy/dx, it may be easier to see this if I write the LHS as e^y(x) where I am showing that y is a function of x, so using the chain rule we have differentiated e^y and then multiplied by the derivative of the inside function. Differentiating the RHS just gives 1 as it is just x. So we now have e^y.dy/dx=1. Re-arranging to give dy/dx=1/(e^y). But remember, e^y=x. Thus dy/dx=1/x. In my explanation I have assumed knowledge of the derivitive of the exponential function, knowledge of the natural logarithm and of the chain rule. If this was not yet present my explanation would be longer and different.

Answered by Milo G. Maths tutor

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