Differentiate y=ln(ln(x)) with respect to x.

To solve this question we need to understand the process of implicit differentiation, which is a case of using the chain rule. If you remember the chain rule states that for y=f(g(x)), we have y'=f'(g(x))g'(x), so that we treat y as being composed of two functions and differentiate them individually, then multiply. So instead if we have f(y)=g(x), then using the same rule on the left hand side but with y, and differentiating both sides we get y'f'(y)=g'(x). Now that this is understood we can solve the question. We are given y=ln(ln(x)) so ey = ln(x). Now differentiate this on both sides: y'ey=1/x. Now we're looking for y' on one side and everything else on the other:y'=1/(xey). We're almost there but there's a problem, we want y' with respect to x so we need the right hand side only with x: fortunately we know ey=lnx, so y'=1/(xln(x)). And we are done. Can you differentiate y = ln(ln(x2)) for me?

MK

Related Maths A Level answers

All answers ▸

Particles P and Q of masses 0.4kg and m kg are joined by a light inextensible string over a smooth pulley. When released Q accelerates downward at 2.45ms^-2. Find m.


A circle with centre C has equation x^2 + y^2 +8x -12y = 12


What is the chain rule?


p(x)=2x^3 + 7x^2 + 2x - 3. (a) Use the factor theorem to prove that x + 3 is a factor of p(x). (b) Simplify the expression (2x^3 + 7x^2 + 2x - 3)/(4x^2-1), x!= +- 0.5